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Question Number 519 by Yugi last updated on 25/Jan/15
$${Find}\:{the}\:{sum}\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{3}{r}+\mathrm{5}\right)\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}=\mathrm{5}\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}\:+\mathrm{8}\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:+\mathrm{11}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:+...\left(\mathrm{3}{n}+\mathrm{5}\right)\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}\: \\ $$$${as}\:{a}\:{simple}\:{function}\:{of}\:{n}. \\ $$
Commented by prakash jain last updated on 22/Jan/15
$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{3}{r}+\mathrm{5}\right)\:^{{n}} {C}_{{r}} \\ $$
Answered by prakash jain last updated on 23/Jan/15
![Σ_(r=0) ^n (3r+5)^n C_r =3Σ_(r=0) ^n r ^n C_r +5Σ_(r=0) ^n ^n C_r =3n∙2^(n−1) +5∙2^n Note 2^n =(1+1)^n =^n C_0 +^n C_1 +...+^n C_n Σ_(r=0) ^n r^n C_r =^n C_1 +2^n C_2 +....+n^n C_n =n+2((n(n−1))/(1∙2))+3((n(n−1)(n−2))/(1∙2∙3))+...+n∙((n(n−1)..1)/(1∙2∙...∙n)) =n[1+(((n−1))/1)+(((n−1)(n−2))/(1∙2))+...+(((n−1)(n−2)...1)/(1∙2∙...(n−1)))] =n[^(n−1) C_0 +^(n−1) C_1 +^(n−1) C_2 +...+^(n−1) C_(n−1) ] =n∙2^(n−1)](Q527.png)
$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{3}{r}+\mathrm{5}\right)\:^{{n}} {C}_{{r}} =\mathrm{3}\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{r}\:\:^{{n}} {C}_{{r}} +\mathrm{5}\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{r}} \\ $$$$=\mathrm{3}{n}\centerdot\mathrm{2}^{{n}−\mathrm{1}} +\mathrm{5}\centerdot\mathrm{2}^{{n}} \\ $$$${Note} \\ $$$$\mathrm{2}^{{n}} =\left(\mathrm{1}+\mathrm{1}\right)^{{n}} =^{{n}} {C}_{\mathrm{0}} +^{{n}} {C}_{\mathrm{1}} +...+^{{n}} {C}_{{n}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{r}\:^{{n}} {C}_{{r}} =\:^{{n}} {C}_{\mathrm{1}} +\mathrm{2}\:^{{n}} {C}_{\mathrm{2}} +....+{n}\:^{{n}} {C}_{{n}} \\ $$$$={n}+\mathrm{2}\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{1}\centerdot\mathrm{2}}+\mathrm{3}\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}}+...+{n}\centerdot\frac{{n}\left({n}−\mathrm{1}\right)..\mathrm{1}}{\mathrm{1}\centerdot\mathrm{2}\centerdot...\centerdot{n}} \\ $$$$={n}\left[\mathrm{1}+\frac{\left({n}−\mathrm{1}\right)}{\mathrm{1}}+\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{1}\centerdot\mathrm{2}}+...+\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)...\mathrm{1}}{\mathrm{1}\centerdot\mathrm{2}\centerdot...\left({n}−\mathrm{1}\right)}\right] \\ $$$$={n}\left[\:^{{n}−\mathrm{1}} {C}_{\mathrm{0}} +\:^{{n}−\mathrm{1}} {C}_{\mathrm{1}} +\:^{{n}−\mathrm{1}} {C}_{\mathrm{2}} +...+\:^{{n}−\mathrm{1}} {C}_{{n}−\mathrm{1}} \right] \\ $$$$={n}\centerdot\mathrm{2}^{{n}−\mathrm{1}} \\ $$
Commented by 112358 last updated on 23/Jan/15
$${Thanks}\:{a}\:{lot} \\ $$