Question

If x⁴ occurs in the t^th term in the expansion of `(x^4 + 1/x^3)^15`, then the value oft is equal to:

पर्याय

• 7

• 8

• 9

• 10

Answer 1

9

Explanation:

The binomial expansion of `(x + a)^n` gives `(t + 1)^(th)` term = `T_(t + 1) = ""^nC_t  x^(n - t)  a^t`

We have expansion of `(x^4 + 1/x^3)^15`

 On comparing with `(x + a)^n`, we get `x = x^4, a = 1/x^3, n` = 15

∴ t^th term = `T_t = ""^15C_(t - 1) (x^4)^(15 - (t - 1)) . (1/x^3)^(t - 1)`

= `""^15C_(t - 1)  (x)^(60 - 4t + 4) . (x)^(- 3t + 3) = ""^15C_(t - 1)  (x)^(67 - 7t)`

Since, `x^4` occurs in the tth term

∴ 67 – 7t = 4

⇒ 7t = 63, t = 9