If the constant term in the binomial expansion of x 2-1- x n is 15 - then n

T_r+1 = ^nC_r(x^2)^n r( 1)^rx^ r = ^nC_r x^2n 3r( 1)^r For the constant term 2n = 3r⇒ r=2n3 Hence, possible values of n=3,6,9⋯ nbsp; and constant term will ...

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Standard XII

Mathematics

Binomial Coefficients

If the consta...

Question

If the constant term in the binomial expansion of ${(x^{2} - \frac{1}{x})}^{n}$ is $15,$ then $n =$

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Solution

$T_{r + 1} =^{n} C_{r} (x^{2})^{n - r} (- 1)^{r} x^{- r}$
$=^{n} C_{r} x^{2 n - 3 r} (- 1)^{r}$
For the constant term $2 n = 3 r \Rightarrow r = \frac{2 n}{3}$
Hence, possible values of $n = 3, 6, 9 \dots$
and constant term will be $^{n} C_{2 n / 3}$
Now putting the value(s) of $n$ in the above expression, we get
$^{3} C_{2} = 3,^{6} C_{4} = 15,^{9} C_{6} = 84 \dots$
Hence, $n = 6$

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Binomial Coefficients

Standard XII Mathematics

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If the constant term in the binomial expansion of (x2−1x)n is 15, then n=

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If the constant term in the binomial expansion of ${(x^{2} - \frac{1}{x})}^{n}$ is $15,$ then $n =$