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$C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots .+C_{n-r} C_n=$
Options:
Solution:
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Verified Answer
The correct answer is:
$\frac{(2 n)!}{(n-r)!(n+r)!}$
$\begin{aligned} & (1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_r x^r+\ldots ..(i)\\ & \left(1+\frac{1}{x}\right)^n=C_0+C_1 \frac{1}{x}+C_2 \frac{1}{x^2}+\ldots .+C_r \frac{1}{x^r}+\ldots(ii)\end{aligned}$
Multiplying both sides and equating coefficient of $x^r$ in $\frac{1}{x^n}(1+x)^{2 n}$ or the coefficient of $x^{n+r}$ in $(1+x)^{2 n}$ we get the value of required expression
$={ }^{2 n} C_{n+r}=\frac{(2 n)!}{(n-r)!(n+r)!}$
Trick : Solving conversely.
Put $n=1$ and $r=0$ in first term, (given condition)
(i) $\left.{ }^1 C_0{ }^1 C_0+{ }^1 C_1{ }^1 C_1=1+1=2 \quad, (r \leq n\right)$
Put $n=2, r=1$, then
(ii) ${ }^2 C_0{ }^2 C_1+{ }^2 C_1{ }^2 C_2=2+2=4$
Now check the options
(a) (i) Put $n=1, r=0$, we get $\frac{2!}{(1)!(1)!}=2$
(ii) Put $n=2, r=1$, we get $\frac{4!}{(1)!(3)!}=4$
Note : Students should remember this question as an identity.
Multiplying both sides and equating coefficient of $x^r$ in $\frac{1}{x^n}(1+x)^{2 n}$ or the coefficient of $x^{n+r}$ in $(1+x)^{2 n}$ we get the value of required expression
$={ }^{2 n} C_{n+r}=\frac{(2 n)!}{(n-r)!(n+r)!}$
Trick : Solving conversely.
Put $n=1$ and $r=0$ in first term, (given condition)
(i) $\left.{ }^1 C_0{ }^1 C_0+{ }^1 C_1{ }^1 C_1=1+1=2 \quad, (r \leq n\right)$
Put $n=2, r=1$, then
(ii) ${ }^2 C_0{ }^2 C_1+{ }^2 C_1{ }^2 C_2=2+2=4$
Now check the options
(a) (i) Put $n=1, r=0$, we get $\frac{2!}{(1)!(1)!}=2$
(ii) Put $n=2, r=1$, we get $\frac{4!}{(1)!(3)!}=4$
Note : Students should remember this question as an identity.
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