Question

The coefficient of x^2 in the expansion(x+4)^3,is​

Answer 1

Answer

(2x​−x23​)10

Tr+1​=10Cr​(2x​)10−r(−x23​)r

         =10Cr​(21​)10−r(−3)r(x10−r)(x−2r)

         =10Cr​(21​)10−r(−3)r[x(10−r)+(−2r)]

⇒(10−r)+(−2r)=4

⇒3r=6

⇒r=2

∴ Coefficent of  x4=10C2​(21​)10−2(−3)2

⇒ 25645×9​ =256405​  =512810​

Hence the answer is 256405​ and 10C2​(21​)89

Answer 2

The coefficient of x^2 in the expansion(x+4)^3,is

(x+3)3 = x3 + 27 + 9x(x+3) =x3+ 27 +9x2+27x. Therefore coefficient of x is 27.