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Question
Answer the following:
The 3rd term of (1 + x)n is 36x2. Find 5th term
Solution
We know that the (r + 1)th term in the expansion of (a + b)n is
tr+1 = nCr, an–r br
Here, a = 1, b = x and for t3, r = 2
∴ 3rd term in (1 + x)n is
t3 = nC2 (1)n-2 x2 = nC2 x2
But 3rd term is given to be 36x2
∴ nC2 x2 =36x2
∴ nC2 = 36
∴ `("n"!)/(2!("n" - 2)!)` = 36
∴ `("n"("n" - 1))/2` = 36
∴ n(n – 1) = 72 = 9 × 8
∴ n(n – 1) = 9(9 – 1)
∴ n = 9
For t5, r = 4
∴ 5th term = t5 = 9C4 (1)9–4 (x)4
= `(9!)/(4!5!) xx x^4`
= `(9 xx 8 xx 7 xx 6)/(1 xx 2 xx 3 xx 4) xx x^4`
= 126x4.
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