The constant term in the expansion of x 2-1- x 216 isA- 16 C 8B- 16 C 7C- 16 C 9D- 16 C 10

The correct option is A ^16C_8 nbsp;(r+1)^th term in the expansion will be ^16C_r (x^2)^r nbsp; ( 1/x^2)^16 r = ^16C_r ( 1)^16 r (x)^4r 32 ⋯ (i) nb ...

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Byju's Answer

Standard XII

Mathematics

Multinomial Expansion

The constant ...

Question

The constant term in the expansion of ${(x^{2} - \frac{1}{x^{2}})}^{16}$ is

^{16} C_{8}

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^{16} C_{7}

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^{16} C_{9}

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^{16} C_{10}

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Solution

The correct option is A $^{16} C_{8}$
$(r + 1)^{t h}$ term in the expansion will be
$^{16} C_{r} (x^{2})^{r} {(\frac{- 1}{x^{2}})}^{16 - r}$

$=^{16} C_{r} (- 1)^{16 - r} (x)^{4 r - 32} \dots (i)$
For constant term, $4 r - 32 = 0$
$\Rightarrow r = 8$
Putting $r = 8 i n (i)$ , we get the constant term as $^{16} C_{8}$ .

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The constant term in the expansion of (x2−1x2)16 is

The correct option is A 16C8 (r+1)th term in the expansion will be 16Cr(x2)r(−1x2)16−r = 16Cr(−1)16−r(x)4r−32 ⋯(i) For constant term, 4r−32=0 ⇒r=8 Putting r=8 in (i), we get the constant term as 16C8.

The constant term in the expansion of ${(x^{2} - \frac{1}{x^{2}})}^{16}$ is