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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))

$${is}\:{it}\:{correct}? \\ $$$${when} \\ $$$${S}_{{n}} =\left\{\right. \\ $$$$\mathrm{1}×\mathrm{2}+\mathrm{1}×\mathrm{3}+\mathrm{1}×\mathrm{4}+\mathrm{1}×\mathrm{5}+.......+\mathrm{1}×{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:+\mathrm{2}×\mathrm{3}+\mathrm{2}×\mathrm{4}+\mathrm{2}×\mathrm{5}+.......+\mathrm{2}×{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}×\mathrm{4}+\mathrm{3}×\mathrm{5}+.......+\mathrm{3}×{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}×\mathrm{5}+.......+\mathrm{4}×{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({n}−\mathrm{1}\right)×{n} \\ $$$$\left.\right\} \\ $$$${find}\:{S}_{{n}} \:. \\ $$$$/////////////// \\ $$$${S}_{{n}} +{S}_{{n}} +\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +...+{n}^{\mathrm{2}} \right)=\left\{\right. \\ $$$$\mathrm{1}×\mathrm{1}+\mathrm{1}×\mathrm{2}+\mathrm{1}×\mathrm{3}+\mathrm{1}×\mathrm{4}+.......+\mathrm{1}×{n}+ \\ $$$$\mathrm{2}×\mathrm{1}+\mathrm{2}×\mathrm{2}+\mathrm{2}×\mathrm{3}+\mathrm{2}×\mathrm{4}+.......+\mathrm{2}×{n}+ \\ $$$$\mathrm{3}×\mathrm{1}+\mathrm{3}×\mathrm{2}+\mathrm{3}×\mathrm{3}+\mathrm{3}×\mathrm{4}+.......+\mathrm{3}×{n}+ \\ $$$$\mathrm{4}×\mathrm{1}+\mathrm{4}×\mathrm{1}+\mathrm{4}×\mathrm{3}+\mathrm{4}×\mathrm{4}+.......+\mathrm{4}×{n}+ \\ $$$$... \\ $$$$\left.{n}×\mathrm{1}+{n}×\mathrm{2}+{n}×\mathrm{3}+{n}×\mathrm{4}+......+{n}×{n}\:\right\} \\ $$$$\Leftrightarrow \\ $$$$=\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+...+{n}\right)\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+...+{n}\right) \\ $$$$\Leftrightarrow \\ $$$$\mathrm{2}{S}_{{n}} +\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}=\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} \\ $$$$\mathrm{2}{S}_{{n}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{3}}\right\} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}×\frac{\mathrm{3}{n}^{\mathrm{2}} −{n}−\mathrm{2}}{\mathrm{6}} \\ $$$${S}_{{n}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}}×\frac{\left(\mathrm{3}{n}+\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{\mathrm{6}} \\ $$$${S}_{{n}} =\frac{\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}{\mathrm{24}} \\ $$$$ \\ $$

Commented by prakash jain last updated on 01/Oct/16

This is correct. I will update my answer. and check for mistakes.

$$\mathrm{This}\:\mathrm{is}\:\mathrm{correct}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{update}\:\mathrm{my}\:\mathrm{answer}. \\ $$$$\mathrm{and}\:\mathrm{check}\:\mathrm{for}\:\mathrm{mistakes}. \\ $$

Commented by prakash jain last updated on 01/Oct/16

Thanks. I have also corrected my answer. I had started wrong. The sum is Σ_(j=1) ^(n−1) [Σ_(i=j+1) ^n j∙i] earlier i had started wrongly Σ_(j=1) ^(n−1) [Σ_(i=j+1) ^n i(i+1)]

$$\mathrm{Thanks}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{also}\:\mathrm{corrected}\:\mathrm{my}\:\mathrm{answer}. \\ $$$$\mathrm{I}\:\mathrm{had}\:\mathrm{started}\:\mathrm{wrong}. \\ $$$$\mathrm{The}\:\mathrm{sum}\:\mathrm{is}\:\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\sum}}{j}\centerdot{i}\right] \\ $$$$\mathrm{earlier}\:\mathrm{i}\:\mathrm{had}\:\mathrm{started}\:\mathrm{wrongly} \\ $$$$\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\sum}}{i}\left({i}+\mathrm{1}\right)\right] \\ $$

Commented by sou1618 last updated on 02/Oct/16

thanks!

$${thanks}! \\ $$

Commented by 314159 last updated on 02/Oct/16

thanks!

$${thanks}! \\ $$

Answered by prakash jain last updated on 02/Oct/16

answer in question

$$\mathrm{answer}\:\mathrm{in}\:\mathrm{question} \\ $$

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