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Question Number 43007 by abdo.msup.com last updated on 06/Sep/18

$$calculate$$$$A_n=\sum _{k=0}^{n}k{C}_n^k$$$$B_n=\sum _{k=0}^n{k}^2{C}_n^k$$$$C_n=\sum _{k=0}^n{k}^3{C}_n^k$$

Commented by maxmathsup by imad last updated on 06/Sep/18

$$letp\left(x\right)=\sum _{k=0}^n{C}_n^k{x}^{k}wehavep\left(x\right)={(x+1)}^{n}and$$$$p^{{}^{\prime }}\left(x\right)=\sum _{k=1}^n{C}_n^{k}k{x}^{k-1}\Rightarrow x{p}^{{}^{\prime }}\left(x\right)=\sum _{k=0}^n{C}_n^{k}k{x}^k\Rightarrow \sum _{k=0}^n{kC}_n^k=p^{\prime }\left(1\right)but$$$$p^{{}^{\prime }}\left(x\right)=n{(x+1)}^{n-1}\Rightarrow p^{{}^{\prime }}\left(1\right)=n{2}^{n-1}=A_{n}alsowehave$$$$p^{{}^{\prime }}\left(x\right)+x{p}^{{}^{″}}\left(x\right)=\sum _{k=1}^n{k}^2{C}_n^k{x}^{k-1}\Rightarrow x{p}^{{}^{\prime }}\left(x\right)+x^2{p}^{{}^{″}}\left(x\right)=\sum _{k=0}^n{k}^2{C}_n^k{x}^k\Rightarrow$$$$\sum _{k=0}^n{k}^2{C}_n^k=p^{{}^{\prime }}\left(1\right)+p^{{}^{″}}\left(1\right)but{p}^{{}^{″}}\left(x\right)=n(n-1){(x+1)}^{n-2}\Rightarrow$$$$p^{″}\left(1\right)=n(n-1)2^{n-2}\Rightarrow B_n=n{2}^{n-1}+n(n-1)2^{n-2}alsowehave$$$$p^{{}^{\prime }}\left(x\right)+{xp}^{\left(2\right)}\left(x\right)+2x{p}^{\left(2\right)}\left(x\right)+x^2{p}^{\left(3\right)}\left(x\right)=\sum _{k=1}^n{k}^3{C}_n^k{x}^{k-1}\Rightarrow$$$${xp}^{{}^{\prime }}\left(x\right)+3x^2{p}^{\left(2\right)}\left(x\right)+x^3{p}^{\left(3\right)}\left(x\right)=\sum _{k=0}^n{k}^3{C}_n^k{x}^k\Rightarrow$$$$\sum _{k=0}^n{k}^3{C}_n^k=p^{{}^{\prime }}\left(1\right)+3p^{\left(2\right)}\left(1\right)+p^{\left(3\right)}\left(1\right)$$$$=n{2}^{n-1}+3n(n-1)2^{n-2}+n(n-1)(n-2)2^{n-3}.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18

$${(1+x)}^n={nC}_0{x}^0+{nC}_{1}x+{nC}_2{x}^2+{nC}_3{x}^3+...+{nC}_n{x}^n$$$$differdntiatew.r.txbothside$$$$n{(1+x)}^{n-1}=0+{nc}_1.1+{nc}_2.2x+{nc}_3.3x^2+..+{nc}_n.{nx}^{n-1}$$$$putx=1$$$$n{\left(2\right)}^{n-1}=0.{nc}_0+1.{nc}_1+2.{nc}_2+3.{nc}_3+..+n.{nc}_n$$$$n.2^{n-1}=\underset{k=0}{\sum ^n}k.{nc}_k\to \mathit{a}\mathit{n}\mathit{s}\to \mathit{A}$$$$n{(1+x)}^{n-1}=0+{nc}_1.1+{nc}_2.2x+{nc}_3.3x^2+..+{nc}_n.{nx}^{n-1}$$$$multiplybothsidebyx$$$$nx{(1+x)}^{n-1}=0.x+{nc}_1.1x+{nc}_2.2x^2+{nc}_3.3x^3+..+{nc}_n.{nx}^n$$$$differentiatebothsidebyx$$$$n(n-1)x{(1+x)}^{n-2}+n{(1+x)}^{n-1}=$$$$0+{nc}_1.1^2+{nc}_2.2^2.x+{nc}_3.3^2.x^2+..+{nc}_n.n^2.x^{n-1}$$$$nowputx=1bothside$$$$n(n-1).2^{n-2}+n.2^{n-1}=0+1^2.{nc}_1+2^2.{nc}_2+3^2.{nc}_3+..+n^2.{nc}_n$$$$n(n-1).2^{n-2}+n.2^{n-1}=\underset{k=0}{\sum ^n}{k}^2.{nc}_k\to \mathit{a}\mathit{n}\mathit{s}\mathit{B}$$$$$$

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