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Question
Answer the following:
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
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