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-

Question Number 8139 by sou1618 last updated on 01/Oct/16

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

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