Question

Find the coefficient of b^6 in the expansion of ((a^2/2) + (2b^2))^10

Answer 1

EXPLANATION.

$\sf \implies \bigg(\dfrac{a^{2} }{2} + 2b^{2} \bigg)^{10}$

As we know that,

Apply binomial expansion in the question, we get.

$\sf \implies ^{10}C_{r} \bigg(\dfrac{a^{2} }{2} \bigg)^{10 - r} . \bigg(2b^{2} \bigg)^{r}$

$\sf \implies ^{10} C_{r} \bigg(\dfrac{a^{2(10 - r)} }{2^{10 - r} } \bigg) . \bigg(2b \bigg)^{2r}$

Coefficient of b⁶ in the expansion, we get.

$\sf \implies b^{2r} = b^{6}$

$\sf \implies 2r = 6.$

$\sf \implies r = 3.$

$\sf \implies T_{r + 1} = 3 + 1 = 4$

Put the value of r = 3 in equation, we get.

$\sf \implies ^{10}C_{3} \bigg( \dfrac{a^{2}^{(10 - 3)} }{2^{(10 - 3)} } \bigg). \bigg(2b \bigg)^{2 \times 3}$

$\sf \implies ^{10}C_{3} \bigg(\dfrac{a^{2(7)} }{2^{(7)} } \bigg) . \bigg(2b \bigg)^{6}$

$\sf \implies ^{10} C_{3} \bigg(\dfrac{a^{14} }{2^{7} } \bigg)$

                                                                                                                       

MORE INFORMATION.

Binomial Theorem with "n" is any index.

$\sf (1 + x)^{n} = 1 + nx + \dfrac{n(n - 1)}{2!} x^{2} + \dfrac{n(n - 1)(n - 2)}{3!} x^{3} + . . . . + \dfrac{n(n - 1). . . (n - r + 1)}{r!} x^{r} + . . . \infty$

$\sf \implies General \ term \ = T_{r + 1} = \dfrac{n(n - 1)(n - 2) . . . . (n - r + 1)}{r!} . x^{r}$

Answer 2

Answer:

The coefficient of x2 in the expansion of (3x−2)7 is −6048