Question 14 Marks
Find the term independent of $x$ in the expansion of $\left(1-x^2\right)\left(x+\frac{2}{x}\right)^6$.
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$2+3.2+4.2^2+\ldots \ldots+(n+1) 2^{n-1}=n \cdot 2^n$
$1^2+4^2+7^2+\ldots \ldots+(3 n-2)^2=\frac{n}{2}\left(6 n^2-3 n-1\right)$
$8+17+26+\ldots . .+(9 n-1)=\frac{n}{2}(9 n+7)$
$\left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)^n=\left(\begin{array}{cc}1 & 2 n \\ 0 & 1\end{array}\right) \forall n \in N$
$(\cos \theta+i \sin \theta)^n=\cos (n \theta)+i \sin (n \theta)$
$5+5^2+5^3+\ldots . .+5^n=\frac{5}{4}\left(5^n-1\right)$
$3^n-2 n-1$ is divisible by 4 .
$\left(2^{4 n}-1\right)$ is divisible by 15 .
$\left(2^{3 n}-1\right)$ is divisible by 7 .
$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots$ to $n$ terms $=\frac{n}{3(2 n+3)}$
$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}$
$1.3+3.5+5.7+\ldots$ to $n$ terms $=\frac{n}{3}\left(4 n^2+6 n-1\right)$
$1.2+2.3+3.4+\ldots+n(n+1)=\frac{n}{3}(n+1)(n+2)$
$1^3+3^3+5^3+\ldots .$. to $n$ terms $=n^2\left(2 n^2-1\right)$
$1^2+3^2+5^2+\ldots .+(2 n-1)^2=\frac{n}{3}(2 n-1)(2 n+1)$
$1^2+2^2+3^2+\ldots . .+n^2=\frac{n(n+1)(2 n+1)}{6}$