Question

Program to find the sum of of the square of natural numbers

Answer 1

Identity:

$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

Proof:

Assume the following:

$1^3+2^3+3^3+...+n^3=1^3+2^3+3^3+...+n^3+(n+1)^3-(n+1)^3$

which implies the following:

$1^3+2^3+3^3+...+n^3=1^3+2^3+3^3+...+n^3+(n+1)^3-(n+1)^3 \\ \\ \Rightarrow\ 0^3+1^3+2^3+3^3+...+n^3=1^3+2^3+3^3+...+n^3+(n+1)^3-(n+1)^3 \\ \\ \Rightarrow\ \displaystyle\sum_{k=0}^{n}k^3=\displaystyle\sum_{k=0}^{n}(k+1)^3-(n+1)^3 \\ \\ \\ \Rightarrow\ \displaystyle\sum_{k=0}^{n}k^3 = \left(\displaystyle\sum_{k=0}^{n}k^3+3k^2+3k+1\right) -(n+1)^3$

$\Rightarrow\ \displaystyle\sum_{k=0}^{n}k^3=\displaystyle\sum_{k=0}^{n}k^3+3\displaystyle\sum_{k=0}^{n}k^2+3\displaystyle\sum_{k=0}^{n}k+\displaystyle\sum_{k=0}^{n}1-(n+1)^3 \\ \\ \\ \Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2+3\displaystyle\sum_{k=0}^{n}k+\displaystyle\sum_{k=0}^{n}1-(n+1)^3=0 \\ \\ \\ \Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=(n+1)^3-3\displaystyle\sum_{k=0}^{n}k-\displaystyle\sum_{k=0}^{n}1$

$\Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=n^3+3n^2+3n+1-3(0+1+2+...+n)-(\underbrace{1+1+1+...1}_\text{n+1}) \\ \\ \\ \Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=n^3+3n^2+3n+1-3 \cdot \frac{n(n+1)}{2}-(n+1) \\ \\ \\ \Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=n^3+3n^2+3n+1-\frac{3n(n+1)}{2}-n-1$

$\Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=\frac{2n^3+6n^2+6n+2-3n(n+1)-2n-2}{2} \\ \\ \\ \Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=\frac{2n^3+6n^2+4n-3n^2-3n}{2} \\ \\ \\ \Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=\frac{2n^3+3n^2+n}{2} \\ \\ \\ \Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=\frac{n(2n^2+3n+1)}{2} \\ \\ \\ \Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=\frac{n(2n^2+2n+n+1)}{2} \\ \\ \\ \Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=\frac{n(2n(n+1)+1(n+1))}{2}$

$\Rightarrow\ 3\displaystyle\sum_{k=0}^{n}k^2=\frac{n(n+1)(2n+1)}{2} \\ \\ \\ \Rightarrow\ \displaystyle\sum_{k = 0}^{n}k^2=\frac{n(n+1)(2n+1)}{2} \div 3 \\ \\ \\ \Rightarrow\ \displaystyle\sum_{k = 0}^{n}k^2=\frac{n(n+1)(2n+1)}{2} \times \frac{1}{3} \\ \\ \\ \Rightarrow\ \displaystyle\sum_{k = 0}^{n}k^2=\frac{n(n+1)(2n+1)}{6} \\ \\ \\ \Rightarrow\ 0^2+1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6} \\ \\ \\ \Rightarrow\ 1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

Hence proved!