Maharashtra State BoardHSC Science (General) 11th
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Answer the following: If the constant term in the expansion of (x3+kx8)11 is 1320, find k - Mathematics and Statistics

Sum

Answer the following:

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

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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
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