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Find the coefficient of x15 in the expansion of (x – x2)10.

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Question

Find the coefficient of x15 in the expansion of (x – x2)10.

Sum
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Solution

The given expression is (x – x2)10

General Term `"T"_(r + 1) = ""^n"C"_r ^(n - r) y^r`

= `""^10"C"_r (x)^(10 - r)  (- x^2)r`

= `""^10"C"_r (x)^(10 - r) (-1)^r * (x^2)^r`

= `(-1)^r * ""^10"C"_r (x)^(10 - r + 2r)`

= `(-1)^r * ""^10"C"_r (x)^(10 + r)`

To find the coefficient of x15

Put 10 + r = 15

⇒ r = 5

∴ Coefficient of x15 = `(-1)^5  ""^10"C"_5`

= `- ""^10"C"_5`

= – 252

Hence, the required coefficient = – 252

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Chapter 8: Binomial Theorem - Exercise [Page 143]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 8 Binomial Theorem
Exercise | Q 6 | Page 143

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