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Question-59639

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Question Number 59639 by Tawa1 last updated on 12/May/19
Commented by Tawa1 last updated on 12/May/19
Please help me to continue. My final answer is wrong. I got: S = ((2n^3 + 3n^2 + n)/6) but answer is S = (n/6) (14n^2 − 9n + 1)
$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{continue}.\:\:\mathrm{My}\:\mathrm{final}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong}. \\ $$$$\:\:\:\mathrm{I}\:\mathrm{got}:\:\:\:\mathrm{S}\:\:=\:\:\frac{\mathrm{2n}^{\mathrm{3}} \:+\:\mathrm{3n}^{\mathrm{2}} \:+\:\mathrm{n}}{\mathrm{6}}\:\:\:\:\:\:\mathrm{but}\:\:\mathrm{answer}\:\mathrm{is}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}\:\:=\:\:\frac{\mathrm{n}}{\mathrm{6}}\:\left(\mathrm{14n}^{\mathrm{2}} \:−\:\mathrm{9n}\:+\:\mathrm{1}\right) \\ $$
Commented by tanmay last updated on 12/May/19
T_r =(r×(n_c_r /n_c_(r−1) ))^2 =(((r×((n!)/(r!(n−r)!)))/((n!)/((n−r+1)!(r−1)!))))^2 =(((r×(r−1)!(n−r+1)!)/(r!(n−r)!)))^2 =(n−r+1)^2 T_r =(n+1)^2 −2(n+1)r+r^2 T_1 =(n+1)^2 −2(n+1)×1+1^2 T_2 =(n+1)^2 −2(n+1)×2+2^2 T_3 =(n+1)^2 −2(n+1)×3+3^2 .... .... T_n =(n+1)^2 −2(n+1)×n+n^2 now add them S=n×(n+1)^2 −2(n+1)(1+2+3+..+n)+(1^2 +2^2 +..+n^2 ) S=n(n+1)^2 −2(n+1)×((n(n+1))/2)+((n(n+1)(2n+1))/6) S=((n(2n^2 +3n+1))/6) S=((2n^3 +3n^2 +n)/6)
$${T}_{{r}} =\left({r}×\frac{{n}_{{c}_{{r}} } }{{n}_{{c}_{{r}−\mathrm{1}} } }\right)^{\mathrm{2}} =\left(\frac{{r}×\frac{{n}!}{{r}!\left({n}−{r}\right)!}}{\frac{{n}!}{\left({n}−{r}+\mathrm{1}\right)!\left({r}−\mathrm{1}\right)!}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{r}×\left({r}−\mathrm{1}\right)!\left({n}−{r}+\mathrm{1}\right)!}{{r}!\left({n}−{r}\right)!}\right)^{\mathrm{2}} \\ $$$$=\left({n}−{r}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${T}_{{r}} =\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({n}+\mathrm{1}\right){r}+{r}^{\mathrm{2}} \\ $$$${T}_{\mathrm{1}} =\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({n}+\mathrm{1}\right)×\mathrm{1}+\mathrm{1}^{\mathrm{2}} \\ $$$${T}_{\mathrm{2}} =\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({n}+\mathrm{1}\right)×\mathrm{2}+\mathrm{2}^{\mathrm{2}} \\ $$$${T}_{\mathrm{3}} =\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({n}+\mathrm{1}\right)×\mathrm{3}+\mathrm{3}^{\mathrm{2}} \\ $$$$…. \\ $$$$…. \\ $$$${T}_{{n}} =\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({n}+\mathrm{1}\right)×{n}+{n}^{\mathrm{2}} \\ $$$${now}\:{add}\:{them} \\ $$$${S}={n}×\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({n}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+..+{n}\right)+\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +..+{n}^{\mathrm{2}} \right) \\ $$$${S}={n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({n}+\mathrm{1}\right)×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${S}=\frac{{n}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${S}=\frac{\mathrm{2}{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{n}}{\mathrm{6}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 13/May/19
That means am right sir. God bless you
$$\mathrm{That}\:\mathrm{means}\:\mathrm{am}\:\mathrm{right}\:\mathrm{sir}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

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