In my earlier article we had derived the formula for sum of cubes of first n natural numbers. Which is given by:
It can be seen from the formula that the sum of cubes of first n natural numbers is equal to the square of nth triangular number. In this article we shall graphically visualise this formula. For this refer figure 1.
Figure 1
If we look at the length and width of figure 1, we are adding numbers from 1 to 4 and then squaring it. Let T represent the total number of balls in figure 1. Then we have
Looking at figure 1 from a different perspective, it is a sum of cubes of numbers from 1 to 4. The red ball is cube of 1, the number of green balls are cube of 2, the yellow balls are cube of 3 and the blue balls are cubes of 4. That is
Thus, we have
Figure 2
Let us analyse this. Refer figure 2. Here, we have the square of (n – 1)th triangular number. Now, let us add n to both the length and the width. Hence, now the total number of balls will be
In that article I had discussed the summation of natural numbers. I am reproducing the relevant portion of that article here.
Let us calculate the sum of first n natural numbers.
Now, let us write this sum in reverse order as given below:
Let us now add both the equations, we get
Hence, the sum of natural numbers from 1 to 100 will be
We can also prove result (i) with visual aids as shown below. Refer figure 1. Here we are adding numbers from 1 to 6. Let s represent the sum. Then we have:
Here Table 1 gives us the Virahanka Sankhayas in the slant diagonals of the Meru Prastaar.
From the above tables it is clear that the Matra Meru is embedded within the Meru Prastaar. Refer Figure 3, here each row of the Matra Meru is coded in the same colour.
The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, … is
There are 8436 steel balls, each with a radius of 1cm, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 balls in the third, 10 in the fourth and so on. Find the number of horizontal layers in the pile?
Interesting and challenging problems, aren’t they? The above two problems were asked in CAT, the entrance exams for admission into India’s most prestigious Business schools “The IIMs”. In these exams you are expected to solve each problem in less than two minutes. These questions are multiple choice questions. But, you get negative marks for the wrong choice.
So, how do you solve such problems not only accurately but also quickly?
Let us begin by solving the first problem.
We have one ‘a‘, two ‘b‘s, three ‘c‘s, four ‘d‘s, and so on. These can be arranged in the shape of a triangle. Hence, it is obvious we are dealing with triangular numbers. For more on triangular numbers refer chapter 11 “How Many Spheres?” from my new book “Meru Prastaar“. Available at https://garudabooks.com/meru-prastaar
Triangular Numbers
We have to find the row number n for the 288th term. Hence, we have
Triangular numbers are given by the formula:
Thus, we have
and
We have,
Thus the 288th term lies in the 24th row. And the 24th character in the English alphabets is ‘x‘. Hence, the 288th term in the given series will be ‘x‘.
Now, let us solve the second problem.
We have 1 ball on top, 3 balls in the second layer, 6 balls in the third, 10 in the fourth and so on. Now 1, 3, 6 and 10 are triangular numbers. Hence, here we have a stack of triangles. And a stack of triangles is a tetrahedron. It is now obvious that 8436 is a tetrahedral number. To understand tetrahedral numbers read chapter 12 “Aryabhata’s Sum of Sums” from my new book “Meru Prastaar“.
Tetrahedral Numbers
The nth tetrahedral number is given by:
Hence, we have
We have
Hence, we have n = 36
Thus, we will have 36 horizontal layers of triangles.
You might be wondering who must have been the first mathematician to calculate the sum of this infinite geometric series?
It was Virasena who had done so in his commentary Dhavala (816 CE) on the Digambara Jain text Shatkhandagama. He had calculated the sum of this infinite series to evaluate the volume of the frustum of a right circular cone.
This summation was proved by Nilakantha Somayaji (1444-1544 CE) (mathematician from Kerala in south India) in Aryabhatiyabhasya his commentary on Aryabhatiya. For details refer my earlier article:
In this article I am going to derive the sum of the given infinite geometric series by geometric means.
Diagram 1
Step 1
Draw an equilateral triangle having an area 1 as shown in Diagram 1.
Diagram 2
Step 2
Join the midpoints of the sides of the equilateral triangle as shown in Diagram 2. This will divide the triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four triangles will be 1/4.
Diagram 3
Step 3
Now join the midpoints of the sides of the top (white) triangle as shown in Diagram 3. This will divide the top triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four new triangles will be 1/4th of ¼.
Now, the sum of the areas of each of the coloured triangles (blue, black and green) will be
Diagram 4
Step 4
Now join the midpoints of the sides of the top (white) triangle as shown in Diagram 4. This will divide the top triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four new triangles will be 1/4th of .
Now, the sum of the areas of each of the coloured triangles (blue, black and green) will be
Diagram 5
Step 5
Now join the midpoints of the sides of the top (white) triangle as shown in Diagram 5. This will divide the top triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four new triangles will be 1/4th of .
Now, the sum of the areas of each of the coloured triangles (blue, black and green) will be
Diagram 6
Step 6
Now join the midpoints of the sides of the top (white) triangle as shown in Diagram 6. This will divide the top triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four new triangles will be 1/4th of .
Now, the sum of the areas of each of the coloured triangles (blue, black and green) will be
Step 7
Repeat the above procedure ad infinitum.
Then, the sum of the areas of each of the coloured triangles (blue, black and green) will be
We can observe from diagram 6 that more the terms you add to the series the closer it gets to 1/3. Hence, we can say that the given series converges to 1/3.
What we popularly call as Fibonacci numbers are predated by Virahanka sankhyas.
In this article we derive a direct combinatorial formula to calculate the Virahanka sankhya.
Refer Table 1. This is left justified Meru Prastaar. If we add up the numbers in the shallow diagonals (represented by the lines in Table 1) we get Virahanka Sankhyanks.
The shallow diagonal that starts from row 0, gives us the 0th Virahanka sankhya, which is 1. The shallow diagonal that starts from row 1, gives us the 1st Virahanka sankhya, which is 1.
The shallow diagonal that starts from row 2, gives us the 2nd Virahanka sankhya. This is = 1 + 1 = 2.
The shallow diagonal that starts from row 3, gives us the 3rd Virahanka sankhya. This is = 1 + 2 = 3.
The shallow diagonal that starts from row 4, gives us the 4th Virahanka sankhya. This is = 1 + 3 + 1 = 5.
The shallow diagonal that starts from row 5, gives us the 5th Virahanka sankhya. This is = 1 + 4 + 3 = 8.
The shallow diagonal that starts from row 6, gives us the 6th Virahanka sankhya. This is = 1 + 5 + 6 +1 = 13.
And so on.
Refer Table 2. This gives the combinatorial formula for the binomial coefficients. Using Table 2, we can calculate the Virahanka sankhyas using the combinatorial formulas. Thus we have:
Do we discern a pattern here? Can we have a formula to calculate the Virahanka sankhyank for a quarter having n matras?
Let us consider a quarter having 5 matras. The metrical forms in a 5 matra prastaar will only have odd numbers of Laghus, i.e. 1, 3 and 5. To put it another way, the metrical forms in a 5 matra prastaar will have 0, 1 and 2 Gurus. The metrical forms in a 5 matra prastaar will have at least 1 Laghu syllable.
The total metrical forms in a 5 matra prastaar are
If there is 1 Guru in the metrical forms then there will be 5 – 1 = 4 syllables in the quarter. If there are 2 Gurus in the metrical forms then there will be 5 – 2 = 3 syllables in the quarter.
In general, for a n matra prastaar, if there are r Gurus in the metrical forms then there will be n – r syllables in the quarter.
Let us now consider a quarter having 6 matras. The metrical forms in a 6 matra prastaar will only have even numbers of Laghus, i.e. 0, 2, 4 and 6. To put it another way, the metrical forms in a 6 matra prastaar will have 0, 1, 2 and 3 Gurus.
The total metrical forms in a 6 matra prastaar are
Let us now consider a quarter having 7 matras. The metrical forms in a 7 matra prastaar will only have odd numbers of Laghus, i.e. 1, 3, 5 and 7. To put it another way, the metrical forms in a 7 matra prastaar will have 0, 1, 2 and 3 Gurus. The metrical forms in a 7 matra prastaar will have at least 1 Laghu syllable.
The total metrical forms in a 7 matra prastaar are
Let us now consider a quarter having 8 matras. The metrical forms in a 8 matra prastaar will only have even numbers of Laghus, i.e. 0, 2, 4, 6 and 8. To put it another way, the metrical forms in a 8 matra prastaar will have 0, 1, 2, 3 and 4 Gurus.
The total metrical forms in a 8 matra prastaar are
From above, we observe that:
In general in a n matra vrutta prastaar, if n is even the metrical forms will have only even number of Laghus. And if n is odd the metrical forms will have only odd number of Laghus.
In general in a n matra vrutta prastaar, if n is even the metrical forms can have maximum n/2 number of Gurus. And if n is odd the metrical forms can have maximum (n – 1)/2 number of Gurus.
Therefore, the total metrical forms in a n matra prastaar are:
If n is even
The number of terms that are added in this formula will be n/2 + 1.
If n is odd
The number of terms that are added in this formula will be (n + 1)/2.
Thus we now have a direct combinatorial formula for calculating the Virahanka sankhyas.
Let us consider the left justified Meru Prastaar as shown in Table 1.
The column 2 contains the sequence of triangular numbers. The nth number of column 2 is the sum of first n numbers of column 1, that is, the sum of first n natural numbers. The 5th number of column 2 is the sum of first 5 numbers (first 5 natural numbers) of column 1, which is 1 + 2 + 3 + 4 + 5 = 15.
The column 3 contains the sequence of tetrahedral numbers. The nth number of column 3 is the sum of first n numbers of column 2, that is, the sum of first n triangular numbers. The 4th number of column 3 is the sum of first 4 numbers (first 4 triangular numbers) of column 2, which is 1 + 3 + 6 + 10 = 20.
The column 4 contains the sequence of pentatope numbers. The nth number of column 4 is the sum of first n numbers of column 3, that is, the sum of first n tetrahedral numbers. The 5th number of column 4 is the sum of first 5 numbers (first 5 tetrahedral numbers) of column 3, which is 1 + 4 + 10 + 20 +35 = 70.
All these sums of sequence of natural numbers are in general called simplex numbers. The numbers in column 1, that is, the sequence of natural numbers are called 1-simplex numbers. The numbers in column 2, that is, the sequence of triangular numbers are called 2-simplex numbers. The numbers in column 3, that is, the sequence of tetrahedral numbers are called 3-simplex numbers. The numbers in column 4, that is, the sequence of pentatope numbers are called 4-simplex numbers. And so on. The numbers in column r are the r-simplex numbers.
The r-simplex number is denoted by Sr. The nth r-simplex number is denoted by Sr(n).
The fifth pentatope number is S4(5) = 70.
Sr(n) is the number in the cell of rth column and (r + n – 1)th row.
The numbers in the cells of Meru Prastaar are binomial coefficients.
Hence, we have
But how do we prove that the nth number of column r is the sum of first n numbers of column (r – 1)?
We have Halayudha’s combinatorial relation:
We have
Also, we have
Substitute this in (1), we have
In the (r – 1)th column of Meru Prastaar the numbers start from (r – 1)th row. Hence, the first n numbers in the (r – 1)th column are from row numbers (r – 1) to (r + n – 2).
This set contains nine members of the Indian cricket team that won the 1983 world cup.
Can you tell me the number of subsets of the above set?
This answer was given by an Ancient Indian Mathematician.
How do we go about counting all the subsets of the above set?
Or for that matter counting all subsets of a set containing n elements?
We know that a set is a subset of itself. Hence, there will be one subset of such a kind. Let T represent the total number of subsets, then T = 1.
Now, a null set will also be a subset of the given set. Hence there will be one subset of such a kind. Hence, now T = 1 + 1 = 2.
Now, I can form subsets by selecting each element of the set. There are 9 elements (or cricketers) in the set. Hence, we will have 9 more subsets. Thus, now T = 2 + 9 = 11.
Now, I can form subsets by selecting 8 out of 9 elements. This is equivalent to skipping each of the 9 terms in turn. This can be done in 9 different ways. Hence, we have 9 more subsets. Thus now, T = 11 + 9 = 20.
Now, I can form subsets by selecting any two of the nine elements. This can be done in
Thus, now T = 20 + 36 = 56.
You must have realised that you can get the total number of subsets by adding up the number of subsets having 0, 1, 2, 3, …, 8, 9 elements from the set.
Thus, we have
These are the binomial coefficients from the 9th row of the Pingala’s Meru Prastaar. Refer my earlier article:
Now, the total of all the binomial coefficients of the 9th row of the Meru Prastaar is T = 2^9 = 512. Thus, the total number of subsets of the given set is 2^9 = 512.
Thus, if now I want to count the total number of subsets of a set containing n elements, I have to sum up the number of subsets containing 0, 1, 2, 3, … upto n elements of the set. That is
Thus, we have
The total number of subsets of a set containing n elements is 2^n.
The device shown in the above videos is a Galton board. It was invented by Sir Francis Galton to demonstrate normal distribution.
You must be wondering how an apparently random phenomenon almost always gives us a similar distribution. Why do we get the type of distribution we get?
Figure 1
Let us analyse in detail what happens inside a Galton’s board.
Refer figure 1. The Galton’s board shown in the figure has 5 rows. Hence, it has 6 bins to collect the balls dropped into the board. The bins are numbered 0 and 1 to 5.
When the ball falls on the only pin in the first row it has equal chance of either going to the left or to the right. When this ball hits any of the pins in the second row again it has equal chance of either going to the left or to the right. Now, when this ball hits any of the pins in the third row again it has equal chance of either going to the left or the right. This happens in every row of the Galton’s board.
If the ball goes to the left of the pins in every row, then the ball falls into the extreme left bin, that is, bin number 0. This path of the ball can be represented as L L L L L, where the leftmost L represents path taken in the first row, the second L represents the path taken in the second row and so on.
Now if the ball goes to the right of the pins in every row, then the ball falls into the extreme right bin, that is, bin number 5. This path can be represented as R R R R R.
In a Galton’s board with 5 rows, a ball can take anyone of the 2^5 = 32 possible paths, from L L L L L to R R R R R. All the paths have equal chance of occurring. A Galton’s board having n rows will have n + 1 bins and a ball in it can take anyone of the 2^n possible paths.
Table 1 enlists all the 32 paths for a 5 row Galton board. How to generate Table 1 can be understood from my earlier article:
Now, let us see in which bin will the ball land when it takes the path given in the second row of Table 1 that is L L L L R. This is as shown in Figure 2.
Figure 2
Now, let us see in which bin will the ball land when it takes the path given in the 5th row of Table 1 that is L L R L L. This is as shown in Figure 3.
Figure 3
From the above discussion we see that all the path combinations that contain only one R will end at bin 1.
Now, let us see in which bin will the ball land when it takes the path given in the 25th row of Table 1 that is R R L L L. This is as shown in Figure 4.
Figure 4
Now, let us see in which bin will the ball land when it takes the path given in the 11th row of Table 1 that is L R L R L. This is as shown in Figure 5.
Figure 5
From the above discussion we see that all the path combinations that contain two Rs will end at bin 2. In the same way all the path combinations that contain 3 Rs will end at bin 3. All the path combinations that contain 4 Rs will end at bin 4. And as already mentioned earlier the path containing all Rs will land in bin 5.
For an n row Galton board, all the path combinations containing ‘k’ Rs will land in the kth bin, where 0 <= k <= n.
Out of all the 32 paths how many paths contain 1, 2, 3, 4 or 5 Rs?
We can count these from Table 1.
Actually these are binomial coefficients from row 5 of the Meru Prastaar. To understand this, refer my article on the Meru Prastaar:
In an n row Galton board, the numbers of paths that contain k Rs are given by
Therefore the probability of the balls to land in bin 0 is
The probability of the balls to land in bin 1 is
The probability of the balls to land in bin 2 is
The graph of the binary coefficients of row 5 of the Meru Prastaar is shown in Figure 6. This is the binary distribution.
Figure 6
If the number of rows n in the Galton board and the sample size (the number of balls put into the board) are sufficiently large then the binomial distribution of the Galton board will approximate the normal distribution.
Numbers in the squares of Meru prastaar are got by adding the numbers in the two squares just above them. This is because of the combinatorial relation given by Halayudha:
There are two ways of understanding this combinatorial relation – Logical and analytical. Let us first understand this in a logical way.
Let’s say that from a group of 14 persons we want to choose a committee of 3 persons. This can be done in ways.
Now let’s say that one more person is added to this group. Hence, there are now 15 persons in this group. We now have to choose 3 persons from this group. This can be done in two ways – either the new entrant is selected for the committee or he is not chosen.
Case 1: the new entrant is not chosen to be in the committee
This is like saying that the new entrant is kept out of the group. Hence, we choose 3 persons from the group of 14 people. This is done in ways.
Case 2: the new entrant is chosen to be in the committee
Hence for the 2 remaining positions in the committee persons will have to be chosen from the original group of 14 people. This can be done in ways.
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