We will derive summation of series with increasing difficulty.

Lets derive the simplest series summation formulas,

\begin{aligned} \sum_{n=1}^{n} {x} = 1 + 2 + 3 + ... + (n-2) + (n-1) + n \end{aligned}

To derive a formula to this series consider adding the same series with all numbers reversed, that is n + (n-1) + (n-2) and so on till 3 + 2 + 1 i.e, check the following equations,

\begin{aligned} & \sum_{n=1}^{n} {x} &&{} = 1 && + 2 && + 3 && + ... && + (n-2) && + (n-1) && + n \\ & \sum_{n=1}^{n} {x} &&{} = n && + (n-1) && + (n-2) && + ... && + 3 && + 2 && + 1 \\ \hline 2 . & \sum_{n=1}^{n} {x} &&{} = (n+1) && + (n+1) && + (n+1) && + ... && + (n+1) && + (n+1) && + (n+1) \end{aligned}

Simplifying, we get

\begin{aligned} 2 . & \sum_{x=1}^{n} {x} &&{} = (n+1) + (n+1) + ... \  n \  times \\ 2 . & \sum_{x=1}^{n}{x} &&{} = n(n+1) \\ & \sum_{x=1}^{n}{x} &&{} = \frac {n(n+1)}{2} \end{aligned}

Well that’s a simple series how about sum of squares! before that we will understand simple properties of \sum symbol.

\begin{aligned} & \sum_{x=1}^{n} (x + 1) &&{} = \sum_{x=1}^{n} x + \sum_{x=1}^{n} 1 = n + \sum_{x=1}^{n} x \\ & \sum_{x=1}^{n} (x + f(x)) &&{} = \sum_{x=1}^{n} x + \sum_{x=1}^{n} f(x) \\ & \sum_{x=1}^{n} (4 + x + 2x^2) &&{} = 4n + \sum_{x=1}^{n} x + 2. \sum_{x=1}^{n} x^2 \end{aligned}

If we observe the formula \sum_{x=1}^{n}{x} = \frac {n(n+1)}{2} , the summation of x is having terms of x^2 and x alone, so if we do \sum \Big( \sum x \Big) we will end up in equation with terms \sum x^2 which can be kept on one side and rest of the terms will be either algebraic function of n or \sum x only.

Lets define a series called S so that,

\begin{aligned} S =& 1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ... + 4) \\ S =& 1.n + 2(n-1) + 3(n-2) + ... + p(n - p + 1) + (p + 1)(n - p) + ... + (n - 1)(n - n + 2) + n(n - n + 1) \\ S =& n + 2n + 3n + ... + pn + (p + 1)n + ... + (n - 1)n + n.n \\ &- [1.0 + 2.1 + 3.2 + ... + p(p - 1) + (p + 1)p + ... + (n - 1)(n - 2) + n(n - 1)] \\ S =& n \sum_{x=1}^{n}{x} - \sum_{x=1}^{n}{x(x - 1)} \end{aligned}

Also the series S expands as follows,

\begin{aligned} S =& \sum_{x=1}^{n} {\Big (\sum_{y=1}^{x} {y} \Big)} =& \frac{1}{2} \sum_{x=1}^{n} {(x^2 + x)} \end{aligned}

Now using above two equations of summation S, we have

\begin{aligned} \frac{1}{2} \sum_{x=1}^{n} {(x^2 + x)} =& n \sum_{x=1}^{n}{x} - \sum_{x=1}^{n}{x(x - 1)} \\ \frac{3}{2} \sum_{x=1}^{n} {x^2} =& \frac{2n + 1}{2} \sum_{x=1}^{n} {x} \\ =& (2n + 1)\frac{n(n + 1)}{4} \\ \sum_{x=1}^{n} {x^2} =& \frac{n(n + 1)(2n + 1)}{6} \end{aligned}

By using this we can now solve sum of series like \sum_{x=1}^{n} (4 + x + 2x^2) mentioned in the above example. Further we can continue with the same technique to get sum of cubic numbers,

\begin{aligned} S =& 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... + (1^2 + 2^2 + 3^2 + ... + 4^2) \\ S =& 1^2.n + 2^2(n-1) + 3^2(n-2) + ... + p^2(n - p + 1) + (p + 1)^2(n - p) + ... + (n - 1)^2(n - n + 2) + n^2(n - n + 1) \\ S =& n + 2^2n + 3^2n + ... + p^2n + (p + 1)^2n + ... + (n - 1)^2n + n^2.n \\ &- [1^2.0 + 2^2.1 + 3^2.2 + ... + p^2(p - 1) + (p + 1)^2p + ... + (n - 1)^2(n - 2) + n^2(n - 1)] \\ S =& n \sum_{x=1}^{n}{x^2} - \sum_{x=1}^{n}{x^2(x - 1)} \\ S =& n \sum_{x=1}^{n}{x^2} - \sum_{x=1}^{n}{(x^3 - x^2)} \\ S =& (n + 1) \sum_{x=1}^{n}{x^2} - \sum_{x=1}^{n}{x^3} \end{aligned}

Also the series S expands as follows,

\begin{aligned} S &= \sum_{x=1}^{n} {\Big (\sum_{y=1}^{x} {y^2} \Big)} \\ &= \frac{1}{6} \sum_{x=1}^{n} {x(x + 1)(2x + 1)} \\ &= \frac{1}{6} \sum_{x=1}^{n} {(2x^3 + 3x^2 + x)} \end{aligned}

Now using above two equations of summation S, we have

\begin{aligned} (n + 1) \sum_{x=1}^{n}{x^2} - \sum_{x=1}^{n}{x^3} &= \frac{1}{6} \sum_{x=1}^{n} {(2x^3 + 3x^2 + x)} \\ \sum_{x=1}^{n} {(2x^3 + 3x^2 + x)} &= 6(n + 1) \sum_{x=1}^{n}{x^2} - 6\sum_{x=1}^{n}{x^3} \\ \sum_{x=1}^{n} {(2x^3 + 6x^3)} &= \sum_{x=1}^{n}{(6(n + 1)x^2 - 3x^2)} - \sum_{x=1}^{n}{x} \\ \sum_{x=1}^{n} {8x^3} &= (6n + 3)\sum_{x=1}^{n}{x^2} - \sum_{x=1}^{n}{x} \\ \sum_{x=1}^{n} {8x^3} &= 3(2n + 1)\frac{n(n + 1)(2n + 1)}{6} - \frac{n(n + 1)}{2} \\ \sum_{x=1}^{n} {x^3} &= \frac{n(n + 1)(2n + 1)^2}{2.8} - \frac{n(n + 1)}{2.8} \\ \sum_{x=1}^{n} {x^3} &= \frac{n(n + 1)(2n + 1)^2 - n(n + 1)}{16} \\ \sum_{x=1}^{n} {x^3} &= \frac{n(n + 1)((2n + 1)^2 - 1)}{16} \\ \sum_{x=1}^{n} {x^3} &= \frac{n(n + 1)(4n^2 + 4n)}{16} \\ \sum_{x=1}^{n} {x^3} &= \frac{n(n + 1)(4n)(n + 1)}{16} \\ \sum_{x=1}^{n} {x^3} &= \frac{n^2(n + 1)^2}{4} \end{aligned}

Hope you will go further to derive higher order formula and who knows a general solution!

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