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