Menu Close

calculate-k-1-n-kC-n-k-3-k-interms-of-n-




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

Leave a Reply

Your email address will not be published. Required fields are marked *