# Algorithm and Summation Formulas and Pattern for Power Series

## Integration from First Principles

Foreword: In this part of the paper, I explain my algorithm and its general formula pattern for power series. (Use the latest Mozilla Firefox Browser, to see the equations properly.)

By: Chrysanthus Date Published: 1 May 2019

### Introduction

In this part of the series, I explain my algorithm and its general formula pattern for power series. Power series are:

$∑ r = 1 n r 0 = 1 0 + 2 0 + 3 0 + … + n 0$ $∑ r = 1 n r 1 = 1 1 + 2 1 + 3 1 + … + n 1$ $∑ r = 1 n r 2 = 1 2 + 2 2 + 3 2 + … + n 2$ $∑ r = 1 n r 3 = 1 3 + 2 3 + 3 3 + … + n 3$ $⋮$ $∑ r = 1 n r m = 1 m + 2 m + 3 m + … + n m$

With my scheme, each series has its own formula. And I have a general formula pattern for all the series.

$The ∑ r = 1 n r 0 Series$ $∑ r = 1 n r 0 = 1 0 + 2 0 + 3 0 + … + n 0$

Using the identity,

$( r + 1 ) 1 - r 1 = 1$

and the binomial theorem, we have:

$∑ r = 1 n r 0 = ( n + 1 ) 1 - 1 1 = n = ∑ r = 1 n 1$

Note that from the Pascal's Triangle, the coefficients of the expansion of the bracketed term are: 1, 1. Since 1/1 is 1, and using the combination symbol, the coefficient of n, the result, may be written as, 1C1/1C1.

$The ∑ r = 1 n r 1 Series$ $∑ r = 1 n r 1 = 1 1 + 2 1 + 3 1 + … + n 1$

Using the identity,

$( r + 1 ) 2 - r 2 = 2 ⁢ r 1 ⁢ 1 1 + 1 2$

and the binomial theorem, we have:

$∑ r = 1 n r 1 = 1 2 ⁢ [ ( n + 1 ) 2 - 1 2 ] - 1 2 ⁢ ∑ r = 1 n r 0$

where the denominator 2, is 2C1.

Substituting for the results in terms of n, and placing in the form of an upper-triangular matrix, gives,

$∑ r = 1 n r 1 = 1 2 ⁢ n 2 + 2 2 ⁢ n 1 - 1 2 ⁢ 1 1 ⁢ n 1$

From Pascal's Triangle and in the expansion of (n+1)2, the coefficients are: 1, 2, 1. In the evaluation of the expression, the last 1 is subtracted out. So the formula can be written, using the combination symbol as:

$∑ r = 1 n r 1 = C 0 2 C 1 2 ⁢ n 2 + C 1 2 C 1 2 ⁢ n 1 - C 2 2 C 1 2 ⁢ C 1 1 C 1 1 ⁢ n 1$

Evaluating the right-hand-side (R.H.S) of the statement gives:

$∑ r = 1 n r 1 = 1 2 ⁢ n 2 + 1 2 ⁢ n 1$
$The ∑ r = 1 n r 2 Series$ $∑ r = 1 n r 2 = 1 2 + 2 2 + 3 2 + … + n 2$

Using the identity,

$( r + 1 ) 3 - r 3 = 3 ⁢ r 2 ⁢ 1 1 + 3 ⁢ r 1 ⁢ 1 2 + 1 3$

and the binomial theorem, we have:

$∑ r = 1 n r 2 = 1 3 ⁢ [ ( n + 1 ) 3 - 1 3 ] - 3 3 ⁢ ∑ r = 1 n r 1 - 1 3 ⁢ ∑ r = 1 n r 0$

where the denominator 3, is 3C1.

Substituting for the results in terms of n, and placing in the form of an upper-triangular matrix, gives,

$∑ r = 1 n r 2 = 1 3 ⁢ n 3 + 3 3 ⁢ n 2 + 3 3 ⁢ n 1 - 3 3 ⁢ 1 2 ⁢ n 2 - 3 3 ⁢ 2 2 ⁢ n 1 + 3 3 ⁢ 1 2 ⁢ 1 1 ⁢ n 1 - 1 3 ⁢ 1 1 ⁢ n 1$

From Pascal's Triangle and in the expansion of (n+1)3, the coefficients are: 1, 3, 3, 1. In the evaluation of the expression, the last 1 is subtracted out. So the formula can be written, using the combination symbol as:

$∑ r = 1 n r 2 = C 0 3 C 1 3 ⁢ n 3 + C 1 3 C 1 3 ⁢ n 2 + C 2 3 C 1 3 ⁢ n 1 - C 2 3 C 1 3 ⁢ C 0 2 C 1 2 ⁢ n 2 - C 2 3 C 1 3 ⁢ C 1 2 C 1 2 ⁢ n 1 + C 2 3 C 1 3 ⁢ C 2 2 C 1 2 ⁢ C 1 1 C 1 1 ⁢ n 1 - C 3 3 C 1 3 ⁢ C 1 1 C 1 1 ⁢ n 1$

Evaluating the R.H.S of the statement gives:

$∑ r = 1 n r 2 = 1 3 ⁢ n 3 + 1 2 ⁢ n 2 + 1 6 ⁢ n 1$
$The ∑ r = 1 n r 3 Series$ $∑ r = 1 n r 3 = 1 3 + 2 3 + 3 3 + … + n 3$

Using the identity,

$( r + 1 ) 4 - r 4 = 4 ⁢ r 3 ⁢ 1 1 + 6 ⁢ r 2 ⁢ 1 2 + 4 ⁢ r 1 ⁢ 1 3 + 1 4$

and the binomial theorem, we have:

$∑ r = 1 n r 3 = 1 4 ⁢ [ ( n + 1 ) 4 - 1 4 ] - 6 4 ⁢ ∑ r = 1 n r 2 - 4 4 ⁢ ∑ r = 1 n r 1 - 1 4 ⁢ ∑ r = 1 n r 0$

where the denominator 4, is 4C1.

Substituting for the results in terms of n, and placing in the form of upper-triangular matrices, gives,

$∑ r = 1 n r 3 = 1 4 ⁢ n 4 + 4 4 ⁢ n 3 + 6 4 ⁢ n 2 + 4 4 ⁢ n 1 - 6 4 ⁢ 1 3 ⁢ n 3 - 6 4 ⁢ 3 3 ⁢ n 2 - 6 4 ⁢ 3 3 ⁢ n 1 + 6 4 ⁢ 3 3 ⁢ 1 2 ⁢ n 2 + 6 4 ⁢ 3 3 ⁢ 2 2 ⁢ n 1 - 6 4 ⁢ 3 3 ⁢ 1 2 ⁢ 1 1 ⁢ n 1 + 6 4 ⁢ 1 3 ⁢ 1 1 ⁢ n 1 - 4 4 ⁢ 1 2 ⁢ n 2 - 4 4 ⁢ 2 2 ⁢ n 1 + 4 4 ⁢ 1 2 ⁢ 1 1 ⁢ n 1 - 1 4 ⁢ 1 1 ⁢ n 1$

From Pascal's Triangle and in the expansion of (n+1)4, the coefficients are: 1, 4, 6, 4, 1. In the evaluation of the expression, the last 1 is subtracted out. So the formula can be written, using the combination symbol as:

$∑ r = 1 n r 3 = C 0 4 C 1 4 ⁢ n 4 + C 1 4 C 1 4 ⁢ n 3 + C 2 4 C 1 4 ⁢ n 2 + C 3 4 C 1 4 ⁢ n 1 - C 2 4 C 1 4 ⁢ C 0 3 C 1 3 ⁢ n 3 - C 2 4 C 1 4 ⁢ C 1 3 C 1 3 ⁢ n 2 - C 2 4 C 1 4 ⁢ C 2 3 C 1 3 ⁢ n 1 + C 2 4 C 1 4 ⁢ C 2 3 C 1 3 ⁢ C 0 2 C 1 2 ⁢ n 2 + C 2 4 C 1 4 ⁢ C 2 3 C 1 3 ⁢ C 1 2 C 1 2 ⁢ n 1 - C 2 4 C 1 4 ⁢ C 2 3 C 1 3 ⁢ C 2 2 C 1 2 ⁢ C 1 1 C 1 1 ⁢ n 1 + C 2 4 C 1 4 ⁢ C 3 3 C 1 3 ⁢ C 1 1 C 1 1 ⁢ n 1 - C 3 4 C 1 4 ⁢ C 0 2 C 1 2 ⁢ n 2 - C 3 4 C 1 4 ⁢ C 1 2 C 1 2 ⁢ n 1 + C 3 4 C 1 4 ⁢ C 2 2 C 1 2 ⁢ C 1 1 C 1 1 ⁢ n 1 - C 4 4 C 1 4 ⁢ C 1 1 C 1 1 ⁢ n 1$

Evaluating the R.H.S of the statement gives:

$∑ r = 1 n r 3 = 1 4 ⁢ n 4 + 1 2 ⁢ n 3 + 1 4 ⁢ n 2 + 0 ⁢ n 1$
$The ∑ r = 1 n r 4 Series$ $∑ r = 1 n r 4 = 1 4 + 2 4 + 3 4 + … + n 4$

Using the identity,

$( r + 1 ) 5 - r 5 = 5 ⁢ r 4 ⁢ 1 1 + 10 ⁢ r 3 ⁢ 1 2 + 10 ⁢ r 2 ⁢ 1 3 + 5 ⁢ r 1 ⁢ 1 4 + 1 5$

and the binomial theorem, we have:

$∑ r = 1 n r 4 = 1 5 ⁢ [ ( n + 1 ) 5 - 1 5 ] - 10 5 ⁢ ∑ r = 1 n r 3 - 10 5 ⁢ ∑ r = 1 n r 2 - 5 5 ⁢ ∑ r = 1 n r 1 - 1 5 ⁢ ∑ r = 1 n r 0$

where the denominator 5, is 5C1.

Substituting for the results in terms of n, and placing in the form of upper-triangular matrices, gives,

$∑ r = 1 n r 4 = 1 5 ⁢ n 5 + 5 5 ⁢ n 4 + 10 5 ⁢ n 3 + 10 5 ⁢ n 2 + 5 5 ⁢ n 1 - 10 5 ⁢ 1 4 ⁢ n 4 - 10 5 ⁢ 4 4 ⁢ n 3 - 10 5 ⁢ 6 4 ⁢ n 2 - 10 5 ⁢ 4 4 ⁢ n 1 - 10 5 ⁢ 6 4 ⁢ 1 3 ⁢ n 3 - 10 5 ⁢ 6 4 ⁢ 3 3 ⁢ n 2 - 10 5 ⁢ 6 4 ⁢ 3 3 ⁢ n 1 + 10 5 ⁢ 6 4 ⁢ 3 3 ⁢ 1 2 ⁢ n 2 + 10 5 ⁢ 6 4 ⁢ 3 3 ⁢ 2 2 ⁢ n 1 - 10 5 ⁢ 6 4 ⁢ 3 3 ⁢ 1 2 ⁢ 1 1 ⁢ n 1 + 10 5 ⁢ 6 4 ⁢ 1 3 ⁢ 1 1 ⁢ n 1 - 10 5 ⁢ 4 4 ⁢ 1 2 ⁢ n 2 - 10 5 ⁢ 4 4 ⁢ 2 2 ⁢ n 1 + 10 5 ⁢ 4 4 ⁢ 1 2 ⁢ 1 1 ⁢ n 1 - 10 5 ⁢ 1 4 ⁢ 1 1 ⁢ n 1 10 5 ⁢ 1 3 ⁢ n 3 + 10 5 ⁢ 3 3 ⁢ n 2 + 10 5 ⁢ 3 3 ⁢ n 1 - 10 5 ⁢ 3 3 ⁢ 1 2 ⁢ n 2 - 10 5 ⁢ 3 3 ⁢ 2 2 ⁢ n 1 + 10 5 ⁢ 3 3 ⁢ 1 2 ⁢ 1 1 ⁢ n 1 - 10 5 ⁢ 1 3 ⁢ 1 1 ⁢ n 1 5 5 ⁢ 1 2 ⁢ n 2 + 5 5 ⁢ 2 2 ⁢ n 1 - 5 5 ⁢ 1 2 ⁢ 1 1 ⁢ n 1 - 1 5 ⁢ 1 1 ⁢ n 1$

From Pascal's Triangle and in the expansion of (n+1)4, the coefficients are: 1, 5, 10, 10, 5, 1. In the evaluation of the expression, the last 1 is subtracted out. So the formula can be written, concisely as:

$∑ r = 1 n r 4 = 1 5 ⁢ n 5 + 1 2 ⁢ n 4 + 1 3 ⁢ n 3 + 0 ⁢ n 2 + - 1 30 ⁢ n 1$
$The ∑ r = 1 n r 5 and ∑ r = 1 n r 6 Series$

The same algorithm and formulas for the power series, when the indices are 5 and 6 are:

Algorithm:

$∑ r = 1 n r 5 = 1 6 ⁢ [ ( n + 1 ) 6 - 1 6 ] - 15 6 ⁢ ∑ r = 1 n r 4 - 20 6 ⁢ ∑ r = 1 n r 3 - 15 6 ⁢ ∑ r = 1 n r 2 - 6 6 ⁢ ∑ r = 1 n r 1 - 1 6 ⁢ ∑ r = 1 n r 0$

Formula:

$∑ r = 1 n r 5 = 1 6 ⁢ n 6 + 1 2 ⁢ n 5 + 5 12 ⁢ n 4 + 0 ⁢ n 3 + - 1 12 ⁢ n 2 + 0 ⁢ n 1$

Algorithm:

$∑ r = 1 n r 6 = 1 7 ⁢ [ ( n + 1 ) 7 - 1 7 ] - 21 7 ⁢ ∑ r = 1 n r 5 - 35 7 ⁢ ∑ r = 1 n r 4 - 35 7 ⁢ ∑ r = 1 n r 3 - 21 7 ⁢ ∑ r = 1 n r 2 - 7 7 ⁢ ∑ r = 1 n r 1 - 1 7 ⁢ ∑ r = 1 n r 0$

Formula:

$∑ r = 1 n r 6 = 1 7 ⁢ n 7 + 1 2 ⁢ n 6 + 1 2 ⁢ n 5 + 0 ⁢ n 4 + - 1 6 ⁢ n 3 + 0 ⁢ n 2 + 1 42 ⁢ n 1$

### Algorithm and Formulas with Patterns

The algorithms and their formulas for series with index 0 to 6 are as follows: $∑ r = 1 n r 0 = ( n + 1 ) 1 - 1 1$ $∑ r = 1 n r 0 = n 1$
$∑ r = 1 n r 1 = 1 2 ⁢ [ ( n + 1 ) 2 - 1 2 ] - 1 2 ⁢ ∑ r = 1 n r 0$ $∑ r = 1 n r 1 = 1 2 ⁢ n 2 + 1 2 ⁢ n 1$
$∑ r = 1 n r 2 = 1 3 ⁢ [ ( n + 1 ) 3 - 1 3 ] - 3 3 ⁢ ∑ r = 1 n r 1 - 1 3 ⁢ ∑ r = 1 n r 0$ $∑ r = 1 n r 2 = 1 3 ⁢ n 3 + 1 2 ⁢ n 2 + 1 6 ⁢ n 1$
$∑ r = 1 n r 3 = 1 4 ⁢ [ ( n + 1 ) 4 - 1 4 ] - 6 4 ⁢ ∑ r = 1 n r 2 - 4 4 ⁢ ∑ r = 1 n r 1 - 1 4 ⁢ ∑ r = 1 n r 0$ $∑ r = 1 n r 3 = 1 4 ⁢ n 4 + 1 2 ⁢ n 3 + 1 4 ⁢ n 2 + 0 ⁢ n 1$
$∑ r = 1 n r 4 = 1 5 ⁢ [ ( n + 1 ) 5 - 1 5 ] - 10 5 ⁢ ∑ r = 1 n r 3 - 10 5 ⁢ ∑ r = 1 n r 2 - 5 5 ⁢ ∑ r = 1 n r 1 - 1 5 ⁢ ∑ r = 1 n r 0$ $∑ r = 1 n r 4 = 1 5 ⁢ n 5 + 1 2 ⁢ n 4 + 1 3 ⁢ n 3 + 0 ⁢ n 2 + - 1 30 ⁢ n 1$
$∑ r = 1 n r 5 = 1 6 ⁢ [ ( n + 1 ) 6 - 1 6 ] - 15 6 ⁢ ∑ r = 1 n r 4 - 20 6 ⁢ ∑ r = 1 n r 3 - 15 6 ⁢ ∑ r = 1 n r 2 - 6 6 ⁢ ∑ r = 1 n r 1 - 1 6 ⁢ ∑ r = 1 n r 0$ $∑ r = 1 n r 5 = 1 6 ⁢ n 6 + 1 2 ⁢ n 5 + 5 12 ⁢ n 4 + 0 ⁢ n 3 + - 1 12 ⁢ n 2 + 0 ⁢ n 1$ $∑ r = 1 n r 6 = 1 7 ⁢ [ ( n + 1 ) 7 - 1 7 ] - 21 7 ⁢ ∑ r = 1 n r 5 - 35 7 ⁢ ∑ r = 1 n r 4 - 35 7 ⁢ ∑ r = 1 n r 3 - 21 7 ⁢ ∑ r = 1 n r 2 - 7 7 ⁢ ∑ r = 1 n r 1 - 1 7 ⁢ ∑ r = 1 n r 0$ $∑ r = 1 n r 6 = 1 7 ⁢ n 7 + 1 2 ⁢ n 6 + 1 2 ⁢ n 5 + 0 ⁢ n 4 + - 1 6 ⁢ n 3 + 0 ⁢ n 2 + 1 42 ⁢ n 1$

Let the index for any series be, m.
Notice that the formula for any series begins with the reciprocal of (m+1), times n raised to the power (m+1), followed by half times n raised to the poewer m, followed by lower powers of n with their coefficients. So the pattern for the general formula for any power series, is:

$∑ r = 1 n r m = 1 m+1 ⁢ n m+1 + 1 2 ⁢ n m + lower powers of n ...$ Pattern for the Summation of Composite Terms

Notice that in the summation of composite terms, there are m+1 columns of terms. The width of each term for a column, grows in size and reduces, generally reducing as you go downward.

### Names for the Algorithm and Patterns

You can call these by my name. You can call the algorithm, Forcha's Algorithm for the Power Series. You can call the formula pattern, Forcha's Formula Pattern for the Power Series. You can call the summation pattern of composite terms, Forcha's Summation Pattern for the Power Series.

That is it for this part of the series. We take a break here and continue in the next part.

Chrys