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\(C^n_n+C^{n-1}_n+C^{n-2}_n=37\)
\(\Leftrightarrow1+\dfrac{n!}{\left(n-1\right)!}+\dfrac{n!}{\left(n-2\right)!2!}=37\)
\(\Leftrightarrow1+n+\dfrac{n\left(n-1\right)}{2}=37\)
\(\Rightarrow n=8\)
\(P=\left(2+5x\right)\left(1-\dfrac{x}{2}\right)^8=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{x}{2}\right)^k\right)\)
\(=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)
\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5x\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)
\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^{k+1}\right)\)
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:\(\left[2.C^3_8\left(-\dfrac{1}{2}\right)^3+5C^2_8\left(-\dfrac{1}{2}\right)^2\right]x^3\)
`2^n C_n ^0+2^[n-1] C_n ^1+2^[n-2] +... +C_n ^n=59049`
`<=>(2+1)^n=59049`
`<=>3^n=59049`
`<=>n=10 =>(2x^2+1/[x^3])^10`
Xét số hạng thứ `k+1:`
`C_10 ^k (2x^2)^[10-k] (1/[x^3])^k ,0 <= k <= 10`
`=C_10 ^k 2^[10-k] x^[20-5k]`
Số hạng chứa `x_5` xảy ra `<=>20-5k=5<=>k=3`
Với `k=3` thì số hạng cần tìm là: `C_10 ^3 2^[10-3] x^5=15360 x^5`
2/ \(\left(a+b\right)^k\Rightarrow k+1\left(so-hang\right)\)
\(\Rightarrow n+6+1=17\Rightarrow n=10\)
6/ \(\left(2a-1\right)^6=\sum\limits^6_{k=0}C^k_6.2^{6-k}.\left(-1\right)^k.a^{6-k}\)
\(\Rightarrow tong-3-so-hang-dau=C^0_6.2^6+C^1_6.2^5.\left(-1\right)+C^2_6.2^4.\left(-1\right)^2=...\)
7/ \(\left(x-\sqrt{y}\right)^{16}=\left(x-y^{\dfrac{1}{2}}\right)^{16}\)
\(\Rightarrow tong-2-so-hang-cuoi=C^{16}_{16}+C^{15}_{16}=...\)
\(C_2^2+C_3^2+...+C_n^2=C_3^3+C_3^2+C_4^2+...+C_n^2\) (do \(C_2^2=C_3^3=1\))
\(=C_4^3+C_4^2+C_5^2+...+C_n^2=C_5^3+C_5^2+...+C_n^2\)
\(=...=C_n^3+C_n^2=C_{n+1}^3\)
Do đó:
\(2C_{n+1}^3=3A_{n+1}^2\Leftrightarrow\dfrac{2.\left(n+1\right)!}{3!.\left(n-2\right)!}=\dfrac{3.\left(n+1\right)!}{\left(n-1\right)!}\)
\(\Leftrightarrow n-1=9\Rightarrow n=10\)
\(\Rightarrow P=\left(1-x-3x^3\right)^{10}=\sum\limits^{10}_{k=0}C_{10}^k\left(-x-3x^3\right)^k\)
\(=\sum\limits^{10}_{k=0}C_{10}^k\left(-1\right)^k\left(x+3x^3\right)^k=\sum\limits^{10}_{k=0}\sum\limits^k_{i=0}C_{10}^kC_k^i\left(-1\right)^kx^i.3^{k-i}.x^{3\left(k-i\right)}\)
\(=\sum\limits^{10}_{k=0}\sum\limits^k_{i=0}C_{10}^kC_k^i\left(-1\right)^k.3^{k-i}.x^{3k-2i}\)
Ta có: \(\left\{{}\begin{matrix}0\le i\le k\le10\\i;k\in N\\3k-2i=4\end{matrix}\right.\) \(\Rightarrow\left(i;k\right)=\left(1;2\right);\left(4;4\right)\)
Hệ số: \(C_{10}^2C_2^1\left(-1\right)^2.3^1+C_{10}^4C_4^4.\left(-1\right)^4.3^0=...\)
\(\Rightarrow he-so:\left[{}\begin{matrix}C^9_{10}C^1_9\left(-3\right)^{10-9}\left(-1\right)=270\\C^{10}_{10}C^4_{10}\left(-3\right)^{10-10}.\left(-1\right)^4=210\end{matrix}\right.\)
Giả thiết tương đương:
\(C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}=2^{100}\) (thay \(1=C_{2n+1}^{2n+1}\))
Mặt khác:
\(C_{2n+1}^{2n+1}=C_{2n+1}^0\)
\(C_{2n+1}^{2n}=C_{2n+1}^1\)
....
\(C_{2n+1}^{n+1}=C_{2n+1}^n\)
Cộng vế:
\(\Rightarrow C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n+1}=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n\)
\(\Rightarrow2\left(C_{2n+1}^{n+1}+...+C_{2n+1}^{2n+1}\right)=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^{2n+1}\)
\(\Rightarrow2.2^{100}=2^{2n+1}\) (đẳng thức cơ bản: \(\sum\limits^n_{k=0}C_n^k=2^n\))
\(\Leftrightarrow2^{101}=2^{2n+1}\)
\(\Rightarrow2n+1=101\)
\(\Rightarrow n=50\)
SHTQ trong khai triển: \(C_{50}^k.\left(x^{-3}\right)^k.\left(x^2\right)^{50-k}=C_{50}^kx^{100-5k}\)
\(100-5k=20\Rightarrow k=16\)
Hệ số: \(C_{50}^{16}\)
