Cocfficient of x^{5} in (x+3)^{8}

Hindi

7.Binomial Theorem

medium

Find the cocfficient of $x^{5}$ in $(x+3)^{8}$

$1512$

Solution

It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right),$ in the binomial expansion of $(a+b)^{n}$ is given by

${T_{r + 1}} = {\,^n}{C_r}{a^{n – r}}{b^r}$

Assuming that $x^{5}$ occurs in the $(r+1)^{t h}$ term of the expansion $(x+3)^{8},$ we obtain

${T_{r + 1}} = {\,^8}{C_r}{(x)^{8 – r}}{(3)^r}$

Comparing the indices of $x$ in $x^{5}$ in $T_{r+1},$

We obtain $r=3$

Thus, the coefficient of $x^{5}$ is ${\,^8}{C_3}{(3)^3} = \frac{{8!}}{{3!5!}} \times {3^3} = \frac{{8 \cdot 7 \cdot 6 \cdot 5!}}{{3 \cdot 2 \cdot 5!}} \cdot {3^3} = 1512$

Standard 11

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Find the cocfficient of $x^{5}$ in $(x+3)^{8}$

Solution

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Show that the middle term in the expansion of $(1+x)^{2 n}$ is
$\frac{1.3 .5 \ldots(2 n-1)}{n !} 2 n\, x^{n},$ where $n$ is a positive integer.