find-the-coefficientof-x-2-in-the-binomial-expansion-of-x-2-2-x-4-

Question Number 54876 by shaddie last updated on 13/Feb/19

find the coefficientof x^{2} in the binomial

expansion of {(x^{2} + \frac{2}{x})}^{4}

Commented by maxmathsup by imad last updated on 13/Feb/19

w e h a v e {(x^{2} + \frac{2}{x})}^{4} = \sum_{k = 0}^{4} C_{4}^{k} {(x^{2})}^{k} {(\frac{2}{x})}^{4 - k}

= \sum_{k = 0}^{4} C_{4}^{k} x^{2 k} 2^{4 - k} x^{k - 4} = \sum_{k = 0}^{4} C_{4}^{k} 2^{4 - k} x^{3 k - 4} t h e c o e f f i c i e n t o f x^{2} i s

λ_{k} / 3 k - 4 = 2 \Rightarrow 3 k = 6 \Rightarrow k = 2 \Rightarrow λ_{k} = λ_{2} = C_{4}^{2} 2^{4 - 2} = 4 C_{4}^{2}

C_{4}^{2} = \frac{4!}{2! 2!} = \frac{4 \times 3}{2} = 6 \Rightarrow λ_{2} = 4 \times 6 = 24 .

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

l e t r + 1 t h t e r m c o n t a i n s x^{2}

4 c_{r} {(x^{2})}^{4 - r} {(\frac{2}{x})}^{r}

4 c_{r} \times x^{8 - 2 r} \times \frac{2^{r}}{x^{r}}

h e n c e x^{8 - 2 r - r} = x^{2}

- 3 r = 2 - 8 \to - 3 r = - 6

r = 2

4 c_{2} \times x^{8 - 3 \times 2} \times 2^{2} \to \frac{4!}{2! 2!} \times 4 \times x^{2}

s o t h e c o e f f i c i e n t i s \to 6 \times 4 = 24