The coefficient of $x^5$ in the expansio of $(1+x)^{21}+(1+x)^{22}+\ldots+(1+x)^{30}$ is
The coefficient of $x^5$ in the expansio of $(1+x)^{21}+(1+x)^{22}+\ldots+(1+x)^{30}$ is
- ${ }^{31} \mathrm{C}_6-{ }^{21} \mathrm{C}_6$
- ${ }^{51} C_5$
- ${ }^9 \mathrm{C}_5$
- ${ }^{30} \mathrm{C}_5+{ }^{20} \mathrm{C}_5$
Solution
As we know, coefficient of $x^r$ in the binomial expansion of $(1+x)^n$ is given by ${ }^n C_r$.
So, coefficient of $x^5$ in the binomial expansion of
$
\begin{aligned}
& (1+x)^{21}+(1+x)^{22}+\ldots .+(1+x)^{30} \\
& ={ }^{21} C_5+{ }^{22} C_5+\ldots .+{ }^{30} C_5 \\
& =\left({ }^{21} C_6+{ }^{21} C_5+{ }^{22} C_5+\ldots .+{ }^{30} C_5\right)-{ }^{21} C_6 \\
& =\left({ }^{22} C_6+{ }^{22} C_5+\ldots .+{ }^{30} C_5\right)-{ }^{21} C_6 \\
& \quad\left[\because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right] \\
& =\left({ }^{23} C_6+{ }^{23} C_6+\ldots .+{ }^{30} C_5\right)-{ }^{21} C_6
\end{aligned}
$
$
=\left({ }^{30} C_6+{ }^{30} C_5\right)-{ }^{21} C_6={ }^{31} C_6-{ }^{21} C_6
$
Asked in: AP EAMCET 2017 (26 Apr Shift 1)
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