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Question Number 63831 by gunawan last updated on 10/Jul/19

If C_r be the coefficient of x^r in (1+x)^n , then the value of Σ_(r=0) ^n (r+1)^2 C_r is

$$\mathrm{If}\:{C}_{{r}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\mathrm{in}\:\left(\mathrm{1}+{x}\right)^{{n}} , \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left({r}+\mathrm{1}\right)^{\mathrm{2}} \:{C}_{{r}} \:\mathrm{is} \\ $$

Answered by mr W last updated on 10/Jul/19

Σ_(r=0) ^n C_r x^r =(1+x)^n Σ_(r=0) ^n C_r x^(r+1) =(1+x)^n x Σ_(r=0) ^n (r+1)C_r x^r =(1+x)^n +n(1+x)^(n−1) x Σ_(r=0) ^n (r+1)C_r x^(r+1) =(1+x)^n x+n(1+x)^(n−1) x^2 Σ_(r=0) ^n (r+1)^2 C_r x^r =(1+x)^n +n(1+x)^(n−1) x+2n(1+x)^(n−1) x+n(n−1)(1+x)^(n−2) x^2 let x=1 Σ_(r=0) ^n (r+1)^2 C_r =2^n +n2^(n−1) +2n2^(n−1) +n(n−1)2^(n−2) =(n+1)2^n +n(n+1)2^(n−2) =(n+1)(n+4)2^(n−2)

$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{r}} {x}^{{r}} =\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{r}} {x}^{{r}+\mathrm{1}} =\left(\mathrm{1}+{x}\right)^{{n}} {x} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left({r}+\mathrm{1}\right){C}_{{r}} {x}^{{r}} =\left(\mathrm{1}+{x}\right)^{{n}} +{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} {x} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left({r}+\mathrm{1}\right){C}_{{r}} {x}^{{r}+\mathrm{1}} =\left(\mathrm{1}+{x}\right)^{{n}} {x}+{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} {x}^{\mathrm{2}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left({r}+\mathrm{1}\right)^{\mathrm{2}} {C}_{{r}} {x}^{{r}} =\left(\mathrm{1}+{x}\right)^{{n}} +{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} {x}+\mathrm{2}{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} {x}+{n}\left({n}−\mathrm{1}\right)\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{2}} {x}^{\mathrm{2}} \\ $$$${let}\:{x}=\mathrm{1} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left({r}+\mathrm{1}\right)^{\mathrm{2}} {C}_{{r}} =\mathrm{2}^{{n}} +{n}\mathrm{2}^{{n}−\mathrm{1}} +\mathrm{2}{n}\mathrm{2}^{{n}−\mathrm{1}} +{n}\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}−\mathrm{2}} \\ $$$$=\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}} +{n}\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}−\mathrm{2}} \\ $$$$=\left({n}+\mathrm{1}\right)\left({n}+\mathrm{4}\right)\mathrm{2}^{{n}−\mathrm{2}} \\ $$

Commented by gunawan last updated on 10/Jul/19

wow thank you Sir

$$\mathrm{wow} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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