The smallest natural number n such that the c | vedclass

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(A) The general term in the expansion of $(x^2 + x^{-3})^n$ is given by $T_{r+1} = ^nC_r (x^2)^{n-r} (x^{-3})^r = ^nC_r x^{2n-2r-3r} = ^nC_r x^{2n-5r}

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(A) The general term in the expansion of $(x^2 + x^{-3})^n$ is given by $T_{r+1} = ^nC_r (x^2)^{n-r} (x^{-3})^r = ^nC_r x^{2n-2r-3r} = ^nC_r x^{2n-5r}$.To find the coefficient of $x$,we set the exponent of $x$ equal to $1$:$2n - 5r = 1 \Rightarrow 2n = 5r + 1$.We are given that the coefficient is $^nC_{23}$,which implies $r = 23$ or $n-r = 23$.Case $1$: If $r = 23$,then $2n = 5(23) + 1 = 115 + 1 = 116$,so $n = 58$.Case $2$: If $n-r = 23$,then $r = n-23$. Substituting this into $2n = 5r + 1$:$2n = 5(n-23) + 1$ $\Rightarrow 2n = 5n - 115 + 1$ $\Rightarrow 3n = 114$ $\Rightarrow n = 38$.Comparing the two possible values for $n$ ($58$ and $38$),the smallest natural number is $38$.

The smallest natural number $n$ such that the coefficient of $x$ in the expansion of $(x^2 + \frac{1}{x^3})^n$ is $^nC_{23}$ is

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