Question

If the constant term in the expansion of `(3x^3 - 2x^2 + 5/x^5)^10` is 2^k. l, where l is an odd integer, then the value of k is equal to ______.

पर्याय

• 6

• 7

• 8

• 9

Answer 1

If the constant term in the expansion of `(3x^3 - 2x^2 + 5/x^5)^10` is 2^k. l, where l is an odd integer, then the value of k is equal to 9.

Explanation:

We have to find the constant term in expansion of `(3x^3 - 2x^2 + 5/x^5)^10`

= `((3x^8 - 2x^7 + 5)/x^5)^10`

= x^–50(3x⁸ – 2x⁷ + 5)¹⁰

For constant term in the expansion of x^–50(3x⁸ – 2x⁷ + 5)¹⁰ we will find the coefficient of x⁵⁰ in (3x⁸ – 2x⁷ + 5)¹⁰

As we know, the coefficient of x^r in the expansion of (a + b + c)ⁿ is given by `(n!)/(r_1!r_2!r_3!) (a)^(r_1)(b)^(r_2)(c)^(r_3)` where r¹ + r² + r³ = n

So here coefficient of x⁵⁰ in the expansion of (3x⁸ – 2x⁷ + 5)¹⁰ 

= `(10!)/(r_1!r_2!r_3!)(3x^8)^(r_1)(-2x^7)^(r_2)(5)^(r_3)`

= `(10!)/(r_1!r_2!r_3!)(3)^(r_1)(-2)^(r_2)(5)^(r_3)(x)^(8r_1 + 7r_2)`

Here r₁ + r₂ + r₃ = 10 and 8r₁ + 7r₂ = 50

Let r₁ = 1 ⇒ r₂ = 6 and r₃ = 3

∴ Coefficient is `(10!)/(1!  6!  3!) (3)^1(-2)^6(5)^3`

= `(10 xx 9 xx 8 xx 7)/(3 xx 2) xx 3 xx (2)^6 xx (5)^3`

= 2 × (5)⁴ × (2)² × 7 × (2)⁶ × (3)²

= (2⁹)(3)²(5)⁴(7)

Now, (2^K)l = (2⁹)(3)²(5)⁴(7)

⇒ K = 9