Find the cocfficient of $x^{5}$ in $(x+3)^{8}$
Find the cocfficient of $x^{5}$ in $(x+3)^{8}$
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 ${\,^8}{C_3}{(3)^3} = \frac{{8!}}{{3!5!}} \times {3^3} = \frac{{8 \cdot 7 \cdot 6 \cdot 5!}}{{3 \cdot 2 \cdot 5!}} \cdot {3^3} = 1512$
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