# In the expansion of (x2-1x2)16, the value of constant term is ______. - Mathematics

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In the expansion of (x^2 - 1/x^2)^16, the value of constant term is ______.

#### Solution

In the expansion of (x^2 - 1/x^2)^16, the value of constant term is 16C8.

Explanation:

Let Tr+1 be the constant term in the expansion of (x^2 - 1/x^2)^16

∴ Tr+1 = ""^16"C"_r (x^2)^(16 - r) ((-1)/x^2)^r

= ""^16"C"_r (x)^(32 - 2r) (-1)^r * 1/x^(2r)

= (-1)^r * ""^16"C"_r (x)^(32 - 2r - 2r)

⇒ (-1)^r * ""^16"C"_r (x)^(32 - 4r)

For getting constant term, 32 – 4r = 0

⇒ r = 8

∴ Tr+1 = (-1)^8 * ""^16"C"_8 = ""^16"C"_8

Concept: General and Middle Terms
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 27 | Page 145

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