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प्रश्न
Find the term containing x6 in the expansion of (2 − x) (3x + 1)9
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उत्तर
(2 − x) (3x + 1)9 = 2(3x + 1)9 − x(3x + 1)9
Consider (3x + 1)9
Here, a = 3x, b = 1, n = 9
We have, tr+1 = nCr an–r .br
= 9Cr (3x)9–r .(1)r
= 9Cr 39–r .x9–r
To get the coefficient of x6 in 2(3x + 1)9, we must have
x9–r = x6
∴ 9 – r = 6
∴ r = 3
Also, to get the coefficient of x6 in x(3x + 1)9, we must have
x.x9–r = x6
∴ x10–r = x6
∴ 10 – r = 6
∴ r = 4
∴ The term containing x6 in the expansion of
2(3x + 1)9 – x(3x + 1)9
=2 9Cr 39–r – 9Cr 39–r
= 2 9C3 39–3 – 9C4 39–4
= 2 × 84 × (36) – 126 × 35
= 2 × 35 [3 × 84 – 63]
= 2 × 243[252 – 63]
= 486 × 189
= 91854
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