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ta có : \(Q=C^1_n+2\dfrac{C_n^2}{C_n^1}+...+k\dfrac{C^k_n}{C_n^{k-1}}+...+n\dfrac{C^n_n}{C_n^{n-1}}\)
\(\Leftrightarrow Q=\dfrac{n!}{1!\left(n-1\right)!}+2\dfrac{1!\left(n-1\right)!}{2!\left(n-2\right)!}+...+k\dfrac{\left(k-1\right)!\left(n-k+1\right)!}{k!\left(n-k\right)!}+...+\dfrac{n\left(n-1\right)!1!}{n!}\)
\(\Leftrightarrow Q=n+\dfrac{2\left(n-1\right)}{2}+...+\dfrac{k\left(n-k+1\right)}{k}+...+\dfrac{n}{n}\)
\(\Leftrightarrow Q=n+\left(n-1\right)+...+\left(n-k+1\right)+...+1\)
\(\Leftrightarrow Q=n^2-\left(1+\left(1+1\right)+\left(1+2\right)+...+\left(n-1\right)\right)\)
1/ \(2C^k_n+5C^{k+1}_n+4C^{k+2}_n+C^{k+3}_n\)
\(=2\left(C^k_n+C_n^{k+1}\right)+3\left(C^{k+1}_n+C^{k+2}_n\right)+\left(C^{k+2}_n+C^{k+3}_n\right)\)
\(=2C_{n+1}^{k+1}+3C_{n+1}^{k+2}+C_{n+1}^{k+3}\)
\(=2\left(C_{n+1}^{k+1}+C_{n+1}^{k+2}\right)+\left(C_{n+1}^{k+2}+C^{k+3}_{n+1}\right)\)
\(=2C_{n+2}^{k+2}+C_{n+2}^{k+3}=C_{n+2}^{k+2}+\left(C_{n+2}^{k+2}+C_{n+2}^{k+3}\right)=C_{n+2}^{k+2}+C_{n+3}^{k+3}\)
Áp dụng ct:C(k)(n)=C(k)(n-1)+C(k-1)(n-1) có:
................C(k-1)(n-1)= C(k)(n) - C(k)(n-1)
tương tự: C(k-1)(n-2)= C(k)(n-1) - C(k)(n-2)
................C(k-1)(n-3)= C(k)(n-2) -C(k)(n-3)
.........................................
................C(k-1)(k-1)= C(k)(k) (=1)
Cộng 2 vế vào với nhau...-> đpcm
\(C^n_n+C^{n-1}_n+C^{n-2}_n=37\)
\(\Leftrightarrow1+\dfrac{n!}{\left(n-1\right)!}+\dfrac{n!}{\left(n-2\right)!2!}=37\)
\(\Leftrightarrow1+n+\dfrac{n\left(n-1\right)}{2}=37\)
\(\Rightarrow n=8\)
\(P=\left(2+5x\right)\left(1-\dfrac{x}{2}\right)^8=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{x}{2}\right)^k\right)\)
\(=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)
\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5x\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)
\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^{k+1}\right)\)
Số hạng chứa \(x^3\) trong \(2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\) là \(2C^3_8.\left(-\dfrac{1}{2}\right)^3x^3\)
Số hạng chứa \(x^3\) trong \(5\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^{k+1}\right)\) là \(5C^2_8.\left(-\dfrac{1}{2}\right)^2x^3\)
Vậy số hạng chứa x3 trong P là:\(\left[2.C^3_8\left(-\dfrac{1}{2}\right)^3+5C^2_8\left(-\dfrac{1}{2}\right)^2\right]x^3\)
\(\frac{P_nC_n^k}{n!A_n^k}=\frac{n!.\frac{n!}{k!\left(n-k\right)!}}{n!.\frac{n!}{\left(n-k\right)!}}=\frac{1}{k!}\)
Chắc là bạn ghi nhầm đề