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Question Number 202774 by BaliramKumar last updated on 03/Jan/24
1. S_n = a(a+d)+(a+d)(a+2d)+......+{a+(n−1)d}{a+(n)d} 2. S_n = a(a+d)(a+2d)+(a+d)(a+2d)(a+3d)+......+{a+(n−1)d}{a+(n)d}{a+(n+1)d}
$$\mathrm{1}.\:\:\:\:\:\:\mathrm{S}_{\mathrm{n}} \:=\:\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)+\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)+……+\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\left(\mathrm{n}\right)\mathrm{d}\right\} \\ $$$$\mathrm{2}.\:\:\:\:\:\:\mathrm{S}_{\mathrm{n}} \:=\:\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)+\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)\left(\mathrm{a}+\mathrm{3d}\right)+……+\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\left(\mathrm{n}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{d}\right\} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 03/Jan/24
Σ_(k=1) ^(k=n) {a+kd)}{(a+(k+1)d} Σ_(k=1) ^(k=n) {a^2 +(2k+1)ad+k(k+1)d^2 } Σ_(k=1) ^n a^2 +Σ_(k=1) (2k+1)ad+Σ_(k=1) ^(k=n) k(k+1)d^2 na^2 +Σ_(k=1) (2k+1)ad+Σ_(k=1) ^(k=n) k(k+1)d^2 na^2 +adΣ_(k=1) ^(k=n) (2k+1)+d^2 Σ_(k=1) ^(k=n) k^2 +k) na^2 +ad{2Σ_(k=1) ^(k=n) k+Σ_(k=1) ^(k=n) 1}+d^2 {Σ_(k=1) ^(k=n) k^2 +Σ_(k=1) ^(k=n) k) na^2 +ad{n+2(((n(n+1))/2))}+d^2 {((n(n+1)(2n+1))/6)+((n(n+1))/2)} na^2 +ad{n^2 +2n}+d^2 {((n(n+1)(2n+1)+3n(n+1))/6)} na^2 +n(n+2)ad+(((n(n+1)(2n+1+3))/6))d^2 na^2 +n(n+2)ad+(((n(n+1)(n+2))/3))d^2
$$\left.\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\sum}}\left\{{a}+{kd}\right)\right\}\left\{\left({a}+\left({k}+\mathrm{1}\right){d}\right\}\right. \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\sum}}\left\{{a}^{\mathrm{2}} +\left(\mathrm{2}{k}+\mathrm{1}\right){ad}+{k}\left({k}+\mathrm{1}\right){d}^{\mathrm{2}} \right\} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}^{\mathrm{2}} +\underset{{k}=\mathrm{1}} {\Sigma}\left(\mathrm{2}{k}+\mathrm{1}\right){ad}+\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\Sigma}}{k}\left({k}+\mathrm{1}\right){d}^{\mathrm{2}} \\ $$$${na}^{\mathrm{2}} +\underset{{k}=\mathrm{1}} {\Sigma}\left(\mathrm{2}{k}+\mathrm{1}\right){ad}+\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\Sigma}}{k}\left({k}+\mathrm{1}\right){d}^{\mathrm{2}} \\ $$$$\left.{na}^{\mathrm{2}} +{ad}\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\Sigma}}\left(\mathrm{2}{k}+\mathrm{1}\right)+{d}^{\mathrm{2}} \underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\Sigma}}{k}^{\mathrm{2}} +{k}\right) \\ $$$${na}^{\mathrm{2}} +{ad}\left\{\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\Sigma}}{k}+\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\Sigma}}\mathrm{1}\right\}+{d}^{\mathrm{2}} \left\{\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\Sigma}}{k}^{\mathrm{2}} +\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\Sigma}}{k}\right) \\ $$$${na}^{\mathrm{2}} +{ad}\left\{{n}+\mathrm{2}\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right)\right\}+{d}^{\mathrm{2}} \left\{\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\} \\ $$$${na}^{\mathrm{2}} +{ad}\left\{{n}^{\mathrm{2}} +\mathrm{2}{n}\right\}+{d}^{\mathrm{2}} \left\{\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{3}{n}\left({n}+\mathrm{1}\right)}{\mathrm{6}}\right\} \\ $$$${na}^{\mathrm{2}} +{n}\left({n}+\mathrm{2}\right){ad}+\left(\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}+\mathrm{3}\right)}{\mathrm{6}}\right){d}^{\mathrm{2}} \\ $$$${na}^{\mathrm{2}} +{n}\left({n}+\mathrm{2}\right){ad}+\left(\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}}\right){d}^{\mathrm{2}} \\ $$
Commented by BaliramKumar last updated on 03/Jan/24
Thanks Sir
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$

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