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$1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+2\cdot6^2+\ldots$

I have tried it a number of ways

First way is I split the series into halves like this $ ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + 4 ^ 2 + 5 ^ 2 + 6 ^ 2 +...) + (2 ^ 2 + 4 ^ 2 + 6 ^ 2 +... ) $

Took last term of first sum as 2n and last term of second sum as n and I am getting something like $ \frac{1}{6}(2n(2n+1)(8n+5)) $

Another way I did this $ 2(1 ^ 2 + 2 ^ 2 + 3 ^ 2 + 4 ^ 2 + 5 ^ 2 + 6 ^ 2 +...) - (1 ^ 2 + 3 ^ 2 + 5 ^ 2 +...) $

And i get $ 4n^2(n+1) $

But my textbook mentions

Sum when last term is even : $ \frac{n(n+1)^2}{2} $

Sum when last term is odd : $ \frac{(n+1)n^2}{2} $

Angelo
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Manu Sm
  • 169

2 Answers2

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We will use the following equality

$\displaystyle\sum\limits_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}6\;.\;\;\color{blue}{(1)}\qquad$ Here is a proof of (1)

If $\,n\,$ is odd ( that is $\,n=2a-1\,,\,$ where $\,a\in\Bbb N$ ) , then

$\begin{align}S_n&=1^2\!+2\!\cdot\!2^2\!+3^2\!+2\!\cdot\!4^2\!+5^2\!+2\!\cdot\!6^2\!+\ldots+2\!\cdot\!(n\!-\!1)^2\!+n^2\!=\\[3pt] &=\sum\limits_{k=1}^nk^2+\!\!\sum\limits_{k=1}^{(n-1)/2}(2k)^2=\sum\limits_{k=1}^nk^2+4\!\!\sum\limits_{k=1}^{(n-1)/2}k^2\!\!\!\underset{\overbrace{\text{ by using (1) }}}{=}\\[3pt] &=\dfrac{n(n+1)(2n+1)}6+4\,\dfrac{\frac{n-1}2\frac{n+1}2n}6=\\[3pt] &=\dfrac{n(n+1)(2n+1)}6+\dfrac{n(n+1)(n-1)}6=\\[3pt] &=\dfrac{n(n+1)3n}6=\dfrac{n^2(n+1)}2\;.\end{align}$

Whereas if $\,n\,$ is even ( that is $\,n=2a\,,\,$ where $\,a\in\Bbb N$ ) , then

$\begin{align}S_n&=1^2\!+2\!\cdot\!2^2\!+3^2\!+2\!\cdot\!4^2\!+5^2\!+2\!\cdot\!6^2\!+\ldots+(n\!-\!1)^2\!+2\!\cdot\!n^2\!=\\[3pt] &=\sum\limits_{k=1}^nk^2+\sum\limits_{k=1}^{n/2}(2k)^2=\sum\limits_{k=1}^nk^2+4\sum\limits_{k=1}^{n/2}k^2\!\!\!\underset{\overbrace{\text{ by using (1) }}}{=}\\[3pt] &=\dfrac{n(n+1)(2n+1)}6+4\,\dfrac{\frac n2\frac{n+2}2(n+1)}6=\\[3pt] &=\dfrac{n(n+1)(2n+1)}6+\dfrac{n(n+1)(n+2)}6=\\[3pt] &=\dfrac{n(n+1)(3n+3)}6=\dfrac{n(n+1)^2}2\;.\end{align}$

Angelo
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3

You need to be careful with the final terms, instead of just writing $\ldots$.

If the last term is $2 \times (2n)^2$, your sum is $$ \sum_{k=1}^{2n} k^2 + \sum_{k=1}^n (2k)^2 $$ but if the last term is $(2n+1)^2$, the sum is $$ \sum_{k=1}^{2n+1} k^2 + \sum_{k=1}^n (2k)^2 $$

Robert Israel
  • 448,999