Question

The observations 1,2,3,...,n are multiplied respectively by 1, 2, 3, ...., n, find the mean.

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Answer 1

$\Large\textrm{Solution}$

$\textbf{- Arithmetic Mean}$

$\boxed{\rm{(A.M.)=\dfrac{\displaystyle\sum^{n}_{i=1}x_{i}}{n}}}$

$\;$

We know,

$\rm{x_{i}=i^{2}}$

$\;$$\;$

Now, let's recall the sum of the first $\rm{n}$ perfect squares.

$\boxed{\rm\sum^{n}_{k=1}k^{2}=\dfrac{n(n+1)(2n+1)}{6}}$

$\;$

Hence,

$\rm{(A.M.)=\dfrac{n(n+1)(2n+1)}{6n}}$

$\;$

$\rm{\therefore(A.M.)=\dfrac{(n+1)(2n+1)}{6}}$

$\;$

$\Large\textrm{Learn More}$

$\textbf{- Formulae}$

$\textbf{\underline{The Sum of the First n Natural Numbers}}$

$\boxed{\rm\displaystyle\sum^{n}_{k=1}k=\dfrac{n(n+1)}{2}}$

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$\textbf{\underline{The Sum of the First n Prefect Squares}}$

Here I will explain how to obtain the sum of the first $\rm n$ perfect squares.

$\;$

We consider a sum of alternating signs.

We know,

$\rm (x+1)^{3}-x^{3}=3x^{2}+3x+1$

$\;$$\boxed{\begin{aligned}2^{3}-1^{3}&=3\cdot1^{2}+3\cdot1+1\\\\3^{3}-2^{3}&=3\cdot2^{2}+3\cdot2+1\\\\4^{3}-3^{3}&=3\cdot3^{2}+3\cdot3+1\\\vdots\\\\(n+1)^{3}-n^{3}&=3n^{2}+3n+1\end{aligned}}$

The sum on LHS: $\rm (n+1)^{3}-1^{3}$

$\;$

The sum on RHS: -

$\;$$\rm\displaystyle3\cdot\sum^{n}_{k=1}k^{2}+3\cdot\sum^{n}_{k=1}k+\sum^{n}_{k=1}1$

$\rm=\displaystyle3\cdot\sum^{n}_{k=1}k^{2}+3\cdot\dfrac{n(n+1)}{2}+n$

$\;$

LHS equals RHS.

$\rm (n+1)^{3}-1^{3}=\displaystyle3\cdot\sum^{n}_{k=1}k^{2}+3\cdot\dfrac{n(n+1)}{2}+n$

$\;$

$\rm\displaystyle3\cdot\sum^{n}_{k=1}k^{2}=n^{3}+3n^{2}+3n-3\cdot\dfrac{n(n+1)}{2}-n$

$\;$

$\rm\displaystyle3\cdot\sum^{n}_{k=1}k^{2}=\dfrac{1}{2}\left\{2n^{3}+6n^{2}+6n-3\cdot n(n+1)-2n\right\}$

$\;$

$\rm\displaystyle3\cdot\sum^{n}_{k=1}k^{2}=\dfrac{1}{2}(2n^{3}+6n^{2}+6n-3n^{2}-3n-2n)$

$\;$$\;$

$\rm\displaystyle3\cdot\sum^{n}_{k=1}k^{2}=\dfrac{1}{2}(2n^{3}+3n^{2}+n)$

$\;$

$\rm\displaystyle3\cdot\sum^{n}_{k=1}k^{2}=\dfrac{n(n+1)(2n+1)}{2}$

$\;$

$\rm\therefore\displaystyle\sum^{n}_{k=1}k^{2}=\dfrac{n(n+1)(2n+1)}{6}$

Hence shown.