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Question Number 178912 by mr W last updated on 22/Oct/22

1×2+2×3+3×4+...+99×100=?

$$\mathrm{1}×\mathrm{2}+\mathrm{2}×\mathrm{3}+\mathrm{3}×\mathrm{4}+...+\mathrm{99}×\mathrm{100}=? \\ $$

Answered by Rasheed.Sindhi last updated on 22/Oct/22

S_n =Σ_(k=1) ^(n) k(k+1) =Σ_(k=1) ^(n) (k^2 +k) =Σ_(k=1) ^(n) k^2 +Σ_(k=1) ^(n) k =((n(n+1)(2n+1))/6)+((n(n+1))/2) =((n(n+1)(2n+1)+3n(n+1))/6) =((n(n+1)(2n+4))/6)=((n(n+1)(n+2))/3) S_(99) =((99(99+1)(99+2))/3)=333300

$$\mathrm{S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}{k}\left({k}+\mathrm{1}\right) \\ $$$$\:\:\:\:=\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}\left({k}^{\mathrm{2}} +{k}\right) \\ $$$$\:\:\:\:=\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}{k}^{\mathrm{2}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}{k} \\ $$$$\:\:\:=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{3}{n}\left({n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\:\:\:=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{4}\right)}{\mathrm{6}}=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}} \\ $$$$\mathrm{S}_{\mathrm{99}} =\frac{\mathrm{99}\left(\mathrm{99}+\mathrm{1}\right)\left(\mathrm{99}+\mathrm{2}\right)}{\mathrm{3}}=\mathrm{333300} \\ $$

Commented by mr W last updated on 22/Oct/22

yes, thanks!

$${yes},\:{thanks}! \\ $$

Commented by mr W last updated on 22/Oct/22

or S_n =Σ_(k=1) ^n k(k+1) S_n =(1/3)Σ_(k=1) ^n k(k+1)(k+2−(k−1)) S_n =(1/3)Σ_(k=1) ^n [k(k+1)(k+2)−(k−1)k(k+1)] S_n =(1/3)[n(n+1)(n+2)−(1−1)1(2+1)] S_n =(1/3)n(n+1)(n+2)

$${or} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\left({k}+\mathrm{1}\right) \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}−\left({k}−\mathrm{1}\right)\right) \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)−\left({k}−\mathrm{1}\right){k}\left({k}+\mathrm{1}\right)\right] \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}}\left[{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)−\left(\mathrm{1}−\mathrm{1}\right)\mathrm{1}\left(\mathrm{2}+\mathrm{1}\right)\right] \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}}{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$

Commented by Rasheed.Sindhi last updated on 23/Oct/22

Grateful sir!

$$\mathbb{G}\boldsymbol{\mathrm{rateful}}\:\boldsymbol{\mathrm{sir}}! \\ $$

Commented by Tawa11 last updated on 23/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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