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प्रश्न
Answer the following:
Find tenth term in the expansion of `(2x^2 + 1/x)^12`
उत्तर
We know that, the (r + 1)th term in the expansion of (a + b)n is
tr+1 = nCr, an-r br
Here a = 2x2, b = `1/x`, n = 12 and for t10, r = 9
∴ tenth term in the expansion of `(2x^2 + 1/x)^12`
=t10 = `""^12"C"_9 (2x^2)^(12 - 9) (1/x)^9`
= `""^12"C"_3 (2x^2)^3 (1/x)^9` ...[∵ nCr = nCn-r]
= `(12 xx 11 xx 10)/(1 xx 2 xx 3) xx 8x^6 xx 1/x^9`
= `1760/x^3`
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