Question

Tìm hệ số của x⁵ trong khai triển:

a) (2+x)⁵(3x-1)⁷

b) (1+x-x²)⁸

Answer 1

Đặt A=(2+x)⁵(3x-1)⁷

^{khai triển ta có:A=(\(_{k=0}^5\Sigma C_5^k2^{5-k}x^k\)}).(\(^7_{i=0}\Sigma C_7^i\left(3x\right)^i\))

=\(\left(_{k=0}^5\Sigma\right)\left(_{i=0}^7\Sigma\right)\left(C_5^kC^i_7\right)\left(x^k.\left(3x\right)^i\right)\)

=số hạng\(\left(C_5^kC^i_7\right)\left(x^k.\left(3x\right)^i\right)\)chứa x⁵ tại k+i=5

có k\(\in\){0,1,2,...5},i\(\in\){0,1,2,...7}

=>(k,i)={(0,5);(1,4);(2,3);(3,2);(4,1);(5,0)}

=>Hệ số của x⁵ là:\(\left(C_5^0C^5_7\right)3^5\)+\(\left(C_5^1C^4_7\right)\left(3^4\right)\)+\(\left(C_5^2C^3_7\right)\left(3^3\right)\)+\(\left(C_5^3C^2_7\right)\left(3^2\right)\)+

\(\left(C_5^4C^1_7\right)\left(3^1\right)\)+\(\left(C_5^5C^0_7\right)3^0\)=30724

Hok tốt!!!

Answer 2

b) ta có (1+x-x²)⁸=(1+(x-x²))⁸

=\(^8_{k=0}\Sigma.C_8^k\left(x-x^2\right)^k\)=\(^8_{k=0}\Sigma.C_8^k\left(x-1\right)^kx^k\)=\(^8_{k=0}\Sigma.C_8^k\left(x-1\right)^kx^k\)