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Let $a_n$ are terms of arithmetic sequence. Derive expression for $\sum_{k=1}^{n} a_k$ using $\sum_{k=1}^{n}a^2_{k+1}-\sum_{k=1}^{n}a^2_k$.


At first, I'm changing index in the first sum to $k=k+1$ so I'm getting $$\sum_{k=2}^{n+1}a^2_{k}-\sum_{k=1}^{n}a^2_k.$$ Then, taking out last term in the first sum and first term in second sum we get $$(n+1)^2 + \sum_{k=2}^n a_k^2 - 1 - \sum_{k=2}^n a_k^2.$$ Which simplifies into $n^2 + 2n$. But we know that $\sum_k^n a_k = \frac{n}{2}(n+1)$. Where do I miss this $\frac{1}{2}$ factor?

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    $\begingroup$ Is $a_k = k$ for all $k$? $\endgroup$ – Umberto P. Sep 28 '16 at 11:19
  • $\begingroup$ It is stated that let $a_n$ are terms of arithmetic sequence. $\endgroup$ – Accelerate to the Infinity Sep 28 '16 at 11:19
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    $\begingroup$ Then you don't generally get $(n+1)^2 - 1$, you get $a_{n+1}^2 - a_1^2$. $\endgroup$ – Daniel Fischer Sep 28 '16 at 11:27
  • $\begingroup$ If $a_n$ is an arithmetic series then $a_{n+1} = a_n + A$ for some constant $A$ so $a_{n+1}^2 = a_n^2 + 2Aa_n + A^2$ thus $\sum_{k=1}^n [a_{k+1}^2 - a_k^2] = nA^2 + 2A\sum_{k=1}^n a_k$. $\endgroup$ – Winther Sep 28 '16 at 11:43
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In your answer, the terms you pulled out of the sum should have been $a_{n+1}^2$ and $a_1^2$, not $(n+1)^2$ and $1$.

Here's another approach: Since $(a_n)_{n \ge 1}$ is an arithmetic sequence, the quantity $d = a_{k+1}-a_k$ is constant for all $k \ge 1$. Hence $$\sum_{k=1}^n a_{k+1}^2 - \sum_{k=1}^n a_k^2 = \sum_{k=1}^n (a_{k+1}-a_k)(a_{k+1}+a_k) = \sum_{k=1}^n d(a_{k+1}+a_k)$$ Now you can play around with reindexing sums to express $\displaystyle\sum_{k=1}^n a_k$ in terms of $\displaystyle\sum_{k=1}^n a_{k+1}^2- \sum_{k=1}^n a_k^2$.

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$$a_{n+1}^2-a_1^2=\sum_{k=1}^{n}a^2_{k+1}-\sum_{k=1}^{n}a^2_k=\sum_{k=1}^{n}\left(a^2_{k+1}-a^2_k\right)=\sum_{k=1}^{n}\left(a_{k+1}-a_k\right)\left(a_{k+1}+a_k\right)$$

$$=\left(a_{k+1}-a_k\right)\sum_{k=1}^{n}\left(a_{k+1}+a_k\right)=\left(a_{k+1}-a_k\right)\left[2\sum_{k=1}^{n}a_k \space -a_1+a_{n+1}\right]$$

where going from line 1 to 2 is justified by the fact that $\left(a_{k+1}-a_k\right)=d$ is constant for arithmetic sequences.

Rearranging, we see:

$$\sum_{k=1}^{n}a_k=\frac{a_{n+1}^2-a_1^2+d(a_1-a_{n+1})}{2d}=\frac{(a_{n+1}-a_{1})(a_{n+1}+a_1-d)}{2d}$$

$$=\frac{a_{n+1}-a_1}{d}\frac{a_1+(a_{n+1}-d)}{2}=n\frac{a_1+a_n}{2}$$

where:

  • first equals sign comes from rearranging
  • second equals sign comes from factoring out $a_{n+1}-a_1$
  • third equals sign comes from splitting the factors
  • fourth equals sign comes from the fact that $a_n$ is an arithmetic sequence, namely that $(a_j-a_k)=(j-k)d$
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