is it correct? when S_n ={ 1×2+1×3+1×4+1×5+.......+1×n +2×3+2×4+2×5+.......+2×n +3×4+3×5+.......+3×n +4×5+.......+4×n .... +(n−1)×n } find S_n . /////////////// S_n +S_n +(1^2 +2^2 +3^2 +4^2 +...+n^2 )={ 1×1+1×2+1×3+1×4+.......+1×n+ 2×1+2×2+2×3+2×4+.......+2×n+ 3×1+3×2+3×3+3×4+.......+3×n+ 4×1+4×1+4×3+4×4+.......+4×n+ ... n×1+n×2+n×3+n×4+......+n×n } ⇔ =(1+2+3+4+...+n)(1+2+3+4+...+n) ⇔ 2S_n +((n(n+1)(2n+1))/6)={((n(n+1))/2)}^2 2S_n =((n(n+1))/2){((n(n+1))/2)−((2n+1)/3)} =((n(n+1))/2)×((3n^2 −n−2)/6) S_n =((n(n+1))/4)×(((3n+2)(n−1))/6) S


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Question Number 8139 by sou1618 last updated on 01/Oct/16
$is it correct? when S_n ={ 1×2+1×3+1×4+1×5+.......+1×n +2×3+2×4+2×5+.......+2×n +3×4+3×5+.......+3×n +4×5+.......+4×n .... +(n−1)×n } find S_n . /////////////// S_n +S_n +(1^2 +2^2 +3^2 +4^2 +...+n^2 )={ 1×1+1×2+1×3+1×4+.......+1×n+ 2×1+2×2+2×3+2×4+.......+2×n+ 3×1+3×2+3×3+3×4+.......+3×n+ 4×1+4×1+4×3+4×4+.......+4×n+ ... n×1+n×2+n×3+n×4+......+n×n } ⇔ =(1+2+3+4+...+n)(1+2+3+4+...+n) ⇔ 2S_n +((n(n+1)(2n+1))/6)={((n(n+1))/2)}^2 2S_n =((n(n+1))/2){((n(n+1))/2)−((2n+1)/3)} =((n(n+1))/2)×((3n^2 −n−2)/6) S_n =((n(n+1))/4)×(((3n+2)(n−1))/6) S_n =(((n−1)n(n+1)(3n+2))/(24))$
$i s i t c o r r e c t ?$ $w h e n$ $S_{n} = {$ $1 \times 2 + 1 \times 3 + 1 \times 4 + 1 \times 5 + . . . . . . . + 1 \times n$ $+ 2 \times 3 + 2 \times 4 + 2 \times 5 + . . . . . . . + 2 \times n$ $+ 3 \times 4 + 3 \times 5 + . . . . . . . + 3 \times n$ $+ 4 \times 5 + . . . . . . . + 4 \times n$ $. . . .$ $+ (n - 1) \times n$ $}$ $f i n d S_{n} .$ $/ / / / / / / / / / / / / / /$ $S_{n} + S_{n} + (1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . + n^{2}) = {$ $1 \times 1 + 1 \times 2 + 1 \times 3 + 1 \times 4 + . . . . . . . + 1 \times n +$ $2 \times 1 + 2 \times 2 + 2 \times 3 + 2 \times 4 + . . . . . . . + 2 \times n +$ $3 \times 1 + 3 \times 2 + 3 \times 3 + 3 \times 4 + . . . . . . . + 3 \times n +$ $4 \times 1 + 4 \times 1 + 4 \times 3 + 4 \times 4 + . . . . . . . + 4 \times n +$ $. . .$ $n \times 1 + n \times 2 + n \times 3 + n \times 4 + . . . . . . + n \times n}$ $\Leftrightarrow$ $= (1 + 2 + 3 + 4 + . . . + n) (1 + 2 + 3 + 4 + . . . + n)$ $\Leftrightarrow$ $2 S_{n} + \frac{n (n + 1) (2 n + 1)}{6} = {\frac{n (n + 1)}{2}}^{2}$ $2 S_{n} = \frac{n (n + 1)}{2} {\frac{n (n + 1)}{2} - \frac{2 n + 1}{3}}$ $= \frac{n (n + 1)}{2} \times \frac{3 n^{2} - n - 2}{6}$ $S_{n} = \frac{n (n + 1)}{4} \times \frac{(3 n + 2) (n - 1)}{6}$ $S_{n} = \frac{(n - 1) n (n + 1) (3 n + 2)}{24}$
Commented by prakash jain last updated on 01/Oct/16

$This is correct . I will update my answer .$ $and check for mistakes .$
Commented by prakash jain last updated on 01/Oct/16

$Thanks . I have also corrected my answer .$ $I had started wrong .$ $The sum is \underset{j = 1}{\sum^{n - 1}} [\underset{i = j + 1}{\sum^{n}} j \cdot i]$ $earlier i had started wrongly$ $\underset{j = 1}{\sum^{n - 1}} [\underset{i = j + 1}{\sum^{n}} i (i + 1)]$
Commented by sou1618 last updated on 02/Oct/16

$t h a n k s!$
Commented by 314159 last updated on 02/Oct/16

$t h a n k s!$
	Answered by prakash jain last updated on 02/Oct/16

	$answer in question$
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