# Answer the following: If the constant term in the expansion of (x3+kx8)11 is 1320, find k - Mathematics and Statistics

Sum

If the constant term in the expansion of (x^3 + "k"/x^8)^11 is 1320, find k

#### Solution

Let tr+1 be the constant term in the expansion of (x^3 + "k"/x^8)^11.

We know that, in the expansion of (a + b)n,

tr+1 = nCr, an–r br

Here a = x3, b = "k"/x^8, n = 11

∴ tr+1 = ""^11"C"_"r" (x^3)^(11 - "r") ("k"/x^8)^"r"

= 11Cr x33–3r · kr · x–8r

= 11Cr kr · x^(33 - 11"r")

But tr+1 is a constant term

∴ power of x = 0

∴ 33 – 11r = 0

∴ r = 3

∴ constant term = 11C3 k3

= (11 xx 10 xx 9)/(1xx 2 xx 3) xx "k"^3 = 165k3

But, the constant term = 1320    ...(Given)

∴ 165k3 = 1320

∴ k3 = 8

∴ k = 2

Concept: General Term in Expansion of (a + b)n
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4 | Q II. (15) | Page 86