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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find the coefficient of $x^5$ in $(x+3)^8$

$\begin{array}{1 1}1512 \\ 1567 \\ 234 \\ 4567 \end{array} $

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  • $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
  • Looking at the pattern of the successive terms we can say that the $(r + 1)\text{th}$ term $T_{r+1} = ^n\large C$$_r\;a^{n–r}b^r$. This is also called the general term of the expansion.
In the expression $(x+3)^8$, assume that $x^5$ occurs in the $(r+1)\text{th}$ term.
$\Rightarrow T_{r+1} = ^8 \large C$$_r\;x^{8–r}3^r$.
Comparing the indices of $x$, we can see that $8-r = 5 \rightarrow r = 3$
$\Rightarrow$ Coefficient of $x^5 = ^8 \large C$$_3 \; 3^3 = 56 \times 27 = 1512$

 

answered Apr 10, 2014 by balaji.thirumalai
edited Apr 10, 2014 by balaji.thirumalai
 

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