find the first three terms in the expansion of [2+x(3+4x)]^5 in ascending power of x - Maths - Binomial Theorem

Question

find the first three terms in the expansion of [2+x(3+4x)]^5 in
ascending power of x

Gursheen kaur. · Accepted Answer

${(2 + 3 x + 4 x^{2})}^{5}$

${(m + n)}^{5} = {}^{5}{C_{0}} m^{5} + {}^{5}{C_{1}} m^{4} (n) + {}^{5}{C_{2}} m^{3} {(n)}^{2} + ..., w h e r e m = 2, n = 3 x + 4 x^{2} = m^{5} + 5 m^{4} n + 10 m^{3} n^{2} + ........... {(2 + 3 x + 4 x^{2})}^{5} = m^{5} + 5 m^{4} n + 10 m^{3} n^{2} + ............. = {(2)}^{5} + 5 {(2)}^{4} (3 x + 4 x^{2}) + 10 {(2)}^{3} {(3 x + 4 x^{2})}^{2} + ........ = 32 + 80 (3 x + 4 x^{2}) + 80 {(3 x + 4 x^{2})}^{2} + ........$

find the first three terms in the expansion of [2+x(3+4x)]^5 in ascending power of x.