Question

1^2+(1^2+2^2)+.......+(1^2+2^2+3^2....+n^2) nbelongs to natural no.​

Answer 1

We have to use these equalities:

$\displaystyle \begin{aligned}1.\ &\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\\ \\ 2.\ &\sum_{i=1}^{n}i^3=\left(\sum_{i=1}^{n}i\right)^2\end{aligned}$

And also,

$\displaystyle \begin{aligned}&\frac{1}{2}\sum_{i=1}^{n}i^2\\ \\ \Longrightarrow\ \ &\frac{1}{2}\cdot \frac{n(n+1)(2n+1)}{6}\\ \\ \Longrightarrow\ \ &\frac{n(n+1)}{2}\cdot \frac{2n+1}{6}\\ \\ \Longrightarrow\ \ &\frac{2n+1}{6}\sum_{i=1}^{n}i\\ \\ \Longrightarrow\ \ &(2n+1)\ \frac{1}{6}\sum_{i=1}^{n}i\end{aligned}$

So,

$\begin{aligned}&(1^2)+(1^2+2^2)+(1^2+2^2+3^2)+...+(1^2+2^2+3^2+...n^2)\\ \\ \Longrightarrow\ \ &\left(\frac{1 \cdot 2 \cdot 3}{6}\right)+\left(\frac{2 \cdot 3 \cdot 5}{6}\right)+\left(\frac{3 \cdot 4 \cdot 7}{6}\right)+...+\left(\frac{n(n+1)(2n+1)}{6}\right)\\ \\ \Longrightarrow\ \ &\sum_{i=1}^{n}\left(\frac{i(i+1)(2i+1)}{6}\right)\\ \\ \Longrightarrow\ \ &\sum_{i=1}^{n}\left(\frac{2i^3+3i^2+i}{6}\right)\end{aligned}$

$\displaystyle \begin{aligned}\\ \\ \Longrightarrow\ \ &\sum_{i=1}^{n}\left(\frac{2i^3}{6}+\frac{3i^2}{6}+\frac{i}{6}\right)\\ \\ \Longrightarrow\ \ &\sum_{i=1}^{n}\left(\frac{i^3}{3}+\frac{i^2}{2}+\frac{i}{6}\right)\\ \\ \Longrightarrow\ \ &\sum_{i=1}^{n}\left(\frac{1}{3}i^3+\frac{1}{2}i^2+\frac{1}{6}i\right)\\ \\ \Longrightarrow\ \ &\frac{1}{3}\sum_{i=1}^{n}i^3+\frac{1}{2}\sum_{i=1}^{n}i^2+\frac{1}{6}\sum_{i=1}^{n}i\end{aligned}$

$\displaystyle \begin{aligned}\\ \\ \Longrightarrow\ \ &\frac{1}{3}\left(\sum_{i=1}^{n}i\right)^2+\frac{1}{2}\sum_{i=1}^{n}i^2+\frac{1}{6}\sum_{i=1}^{n}i\\ \\ \Longrightarrow\ \ &\frac{1}{2}\sum_{i=1}^{n}i^2+\frac{1}{6}\sum_{i=1}^{n}i+\frac{1}{3}\left(\sum_{i=1}^{n}i\right)^2\\ \\ \Longrightarrow\ \ &\left((2n+1)+1+2\sum_{i=1}^{n}i\right)\frac{1}{6}\sum_{i=1}^{n}i\end{aligned}$

$\displaystyle \begin{aligned}\\ \\ \Longrightarrow\ \ &\left(2n+2+2\sum_{i=1}^{n}i\right)\frac{1}{6}\sum_{i=1}^{n}i\\ \\ \Longrightarrow\ \ &\left(n+1+\sum_{i=1}^{n}i\right)\frac{1}{3}\sum_{i=1}^{n}i\\ \\ \Longrightarrow\ \ &\left(n+1+\frac{n(n+1)}{2}\right)\frac{1}{3} \cdot \frac{n(n+1)}{2}\\ \\ \Longrightarrow\ \ &\left(\frac{2(n+1)}{2}+\frac{n(n+1)}{2}\right)\frac{n(n+1)}{6}\\ \\ \Longrightarrow\ \ &\left(\frac{(n+1)(n+2)}{2}\right)\frac{n(n+1)}{6}\end{aligned}$

$\displaystyle \begin{aligned}\Longrightarrow\ \ &\Large \frac{n(n+1)^2(n+2)}{12}\\ \\ \Longrightarrow\ \ &\Large \frac{n(n+1)(n+1)(n+2)}{12}\end{aligned}$

As examples,

$\displaystyle \begin{aligned}1.&\ \ (1^2)&=&\ \ \frac{1 \cdot 2 \cdot 2 \cdot 3}{12}&=&\ \ 1\\ \\ 2.&\ \ (1^2)+(1^2+2^2)&=&\ \ \frac{2\cdot 3\cdot 3\cdot 4}{12}&=&\ \ 6\\ \\ 3.&\ \ (1^2)+(1^2+2^2)+(1^2+2^2+3^2)&=&\ \ \frac{3\cdot 4\cdot 4\cdot 5}{12}&=&\ \ 20\end{aligned}$