i-Find-the-first-three-terms-in-the-expansion-of-2-x-6-in-ascending-power-of-x-ii-Find-the-value-of-k-for-which-there-is-no-term-in-x-2-in-the-expansion-1-kx-2-x-6-

Question Number 21732 by tawa tawa last updated on 02/Oct/17

(i) Find the first three terms in the expansion of {(2 - x)}^{6} in ascending power

of x .

(ii) Find the value of k for which there is no term in x^{2} in the expansion

(1 + kx) {(2 - x)}^{6}

Commented by Tikufly last updated on 02/Oct/17

(i) {(2 - x)}^{6} =^{6} C_{0} 2^{6} -^{6} C_{1} 2^{5} x +^{6} C_{2} 2^{4} x^{2} -^{6} C_{3} 2^{3} x^{3} + \dots .

= 64 - 192 x + {240 x}^{2} - {160 x}^{3} \dots . .

Hence, the required terms are : -

- 192 x + {240 x}^{3} - {160 x}^{3}

(ii) (1 + kx) {(2 - x)}^{6}

= (1 + kx) (64 - 192 x + {240 x}^{2} - {160 x}^{3} \dots \dots .)

Since there is no term of x^{2}

Coefficient of x^{2} = 0

=> 240 - 192 k = 0

=> k = \frac{240}{192}

=> k = \frac{5}{4}

Commented by tawa tawa last updated on 02/Oct/17

God bless you sir .