Đề bài: Chứng minh rằng: $n.4^{n-1}C^0_n-(n-1)4^{n-2}C^1_n+(n-2)4^{n-1}C^2_n-…+(-1)^{n-1}C^{n-1}_n$ $ = C^1_n + 4C^2_n+…+n.2^{n-1}C^n_n, \forall n \in N$
Lời giải
Xét $VT=n.4^{n-1}.C^{0}_{n}-(n-1).4^{n-2}C^{1}_{n}+(n-2).4^{n-3}C^{2}_{n}+…+(-1)^{n-1}.C^{n-1}_{n} $
$SHTQ=(-1)^k(n-k).4^{n-(k+1)}.C^{k}_{n} $
$=(n-k)(-1)^k.4^{n-(k+1)}. \frac{n!}{k!(n-k)!} $
$=(-1)^k.n.4^{n-(k+1)}.\frac{(n-1)!}{k!(n-k-1)!} $
$=(-1)^k.n. C^{k}_{n}.4^{n-(k+1)} $
$VT= n \left ( 4^{n-1}.C^{0}_{n-1} – 4^{n-2}C^{1}_{n-1}+4^{n-3}.C^{2}_{n-1} +…+ (-1)^{n-1}.C^{n-1}_{n-1} \right ) $
$=n \left ( 4-1 \right ){n-1}=n.3^{n-1}(*) $
Xét $VP=C^{1}_{n}+4 C^{2}_{n}+…+n.2^{n-1}.C^{n}_{n} $
$SHTQ= k.2^{k-1}.C^{k}_{n}=k.2^{n-1}\frac{n!}{k!(n-k)!} $
$=\frac{n.2^{k-1}.(n-1)!}{(k-1)!(n-k)!} $
$VP=n \left ( C^{0}_{n-1} +2 C^{1}_{n-1} +…+2^{n-1}. C^{n-1}_{n-1} \right ) $
$=n \left ( 1+2 \right )^{n-1}=n.3^{n-1} (**)$
Từ $(*)$ và $(**)$ ta được đpcm.
Trả lời