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Find the $n^{th}$ term and sum to $n$ terms of the following series. $$1.3+2.4+3.5+……$$

My Attempt:

Here, $n^{th}$ term of $1+2+3+……=n$

$n^{th}$ term of $3+4+5+……=n+2$

Thus,

$n^{th}$ term of the series $1.3+2.4+3.5+……=n(n+2)$

$$t_n=n^2+2n$$

If $S_n$ be the sum to $n$ terms of the series then $$S_n=\sum t_n$$ $$=\sum (n^2+2n)$$

How do I proceed?

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  • $\begingroup$ Is that multiplication or decimals? Also did you mean $n(n+2)$? $\endgroup$ – Andrew Li Mar 20 '18 at 1:26
  • $\begingroup$ Use \cdot e.g $1\cdot 3$ for multiplication. $\endgroup$ – dssknj Mar 20 '18 at 3:20
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\begin{align}\sum_{i=1}^n (i^2+2i) &=\sum_{i=1}^n i^2 + 2\sum_{i=1}^n i\\ &= \frac{n(n+1)(2n+1)}{6}+2\cdot \frac{n(n+1)}{2} \end{align}

You might want to factorize the terms to simplify things.

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With the summation:

$$\sum n^2+2n$$

Split it into two via linearity:

$$\sum n^2 + 2\sum n$$

Then apply for the formulae for the sum of first $n$ numbers and $n$ squares:

$$\sum n^2 = {n(n+1)(2n+1)\over 6}$$ $$\sum n = {n(n+1)\over 2}$$

Thus producing:

$${n(n+1)(2n+1)\over 6} + n(n+1)$$

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