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Question Number 64651 by mathmax by abdo last updated on 20/Jul/19

calculate Σ_(k=1) ^n kC_n ^k 3^k interms of n

$${calculate}\:\sum_{{k}=\mathrm{1}} ^{{n}} {kC}_{{n}} ^{{k}} \:\mathrm{3}^{{k}} \:\:{interms}\:{of}\:{n} \\ $$

Commented by mathmax by abdo last updated on 20/Jul/19

let p(x) =Σ_(k=0) ^n C_n ^k x^k we have p(x)=(x+1)^n and p^′ (x) =Σ_(k=1) ^n k C_n ^k x^(k−1) ⇒ xp^′ (x) =Σ_(k=1) ^n k C_n ^k x^k x=3 ⇒3p^′ (3) =Σ_(k=1) ^n k C_n ^k 3^k but p^′ (x)=n(x+1)^(n−1) ⇒ p^′ (3) =n4^(n−1) ⇒Σ_(k=1) ^n kC_n ^k 3^k =3n 4^(n−1) .

$${let}\:{p}\left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\:\:\:\:\:{we}\:{have}\:{p}\left({x}\right)=\left({x}+\mathrm{1}\right)^{{n}} \:\:\:{and} \\ $$$${p}^{'} \left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{C}_{{n}} ^{{k}} \:{x}^{{k}−\mathrm{1}} \:\Rightarrow\:{xp}^{'} \left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{C}_{{n}} ^{{k}} \:{x}^{{k}} \\ $$$${x}=\mathrm{3}\:\Rightarrow\mathrm{3}{p}^{'} \left(\mathrm{3}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{C}_{{n}} ^{{k}} \:\mathrm{3}^{{k}} \:\:\:\:\:\:{but}\:{p}^{'} \left({x}\right)={n}\left({x}+\mathrm{1}\right)^{{n}−\mathrm{1}} \:\Rightarrow \\ $$$${p}^{'} \left(\mathrm{3}\right)\:={n}\mathrm{4}^{{n}−\mathrm{1}} \:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:{kC}_{{n}} ^{{k}} \:\mathrm{3}^{{k}} \:=\mathrm{3}{n}\:\mathrm{4}^{{n}−\mathrm{1}} \:. \\ $$

Answered by mr W last updated on 20/Jul/19

(1+x)^n =Σ_(k=0) ^n C_n ^k x^k n(1+x)^(n−1) =Σ_(k=0) ^n kC_n ^k x^(k−1) n(1+x)^(n−1) x=Σ_(k=0) ^n kC_n ^k x^k with x=3: n(1+3)^(n−1) 3=Σ_(k=0) ^n kC_n ^k 3^k =Σ_(k=1) ^n kC_n ^k 3^k ⇒Σ_(k=1) ^n kC_n ^k 3^k =3n4^(n−1)

$$\left(\mathrm{1}+{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{k}} \\ $$$${n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{kC}_{{n}} ^{{k}} {x}^{{k}−\mathrm{1}} \\ $$$${n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} {x}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{kC}_{{n}} ^{{k}} {x}^{{k}} \\ $$$${with}\:{x}=\mathrm{3}: \\ $$$${n}\left(\mathrm{1}+\mathrm{3}\right)^{{n}−\mathrm{1}} \mathrm{3}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{kC}_{{n}} ^{{k}} \mathrm{3}^{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{kC}_{{n}} ^{{k}} \mathrm{3}^{{k}} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{kC}_{{n}} ^{{k}} \mathrm{3}^{{k}} =\mathrm{3}{n}\mathrm{4}^{{n}−\mathrm{1}} \\ $$

Commented by mathmax by abdo last updated on 20/Jul/19

thank you sir mrw

$${thank}\:{you}\:{sir}\:{mrw} \\ $$

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