Question

If the constant term in the expansion of$(3x^3−2x^2+\frac{5}{x^5})^{10}$ is 2 k · l, where l is an odd integer, then the value of k is equal to

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

Answer: (D)

9

Answer 2

The correct option is(D): 9

$(3x^3−2x^2+\frac{5}{x^5})^{10}→x^0$

⇒ (3 x 8 – 2 x 7 + 5)10→ x 50

General term of (3 x 8 – 2 x 7 + 5)10 is

$\frac{10!}{p!q!r!}(3x^8)^p(−2x^7)^9(5)^r$

Here 8 p + 7 q = 50 and p + q +r = 10

⇒ p = 1, q = 6, r = 3

$∴\frac{10!}{1!6!r!}3^1 2^6⋅5^3=2^k⋅l$

$⇒5⋅3⋅7⋅53⋅3⋅29=2^kl$

∴ k = 9