Binomial Theorem 2 Question 16
19.
Prove that $C_{1}^{2}-2 \cdot C_{2}^{2}+3 \cdot C_{3}^{2}-\ldots-2 n \cdot C_{2 n}^{2}=(-1)^{n} n \cdot C_{n}$
(1979, 4M)
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Solution:
- We know that, $(1+x)^{2 n}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{2 n} x^{2 n}$
On differentiating both sides w.r.t. $x$, we get $2 n(1+x)^{2 n-1}=C_{1}+2 \cdot C_{2} x+3 \cdot C_{3} x^{2}$
$ +\ldots+2 n C_{2 n} x^{2 n-1} $ $\quad$ …….(i)
and $1-\frac{1}{x}^{2 n}=C_{0}-C_{1} \cdot \frac{1}{x}+C_{2} \cdot \frac{1}{x^{2}}-C_{3} \cdot \frac{1}{x^{3}}$
$ +\ldots+C_{2 n} \cdot \frac{1}{x^{2 n}} $ $\quad$ …….(ii)
On multiplying Eqs. (i) and (ii), we get
$2 n(1+x)^{2 n-1} (1-\frac{1}{x})^{2 n} $
$=\left[C_{1}+2\right. \left.\cdot C_{2} x+3 \cdot C_{3} x^{2}+\ldots+2 n \cdot C_{2 n} x^{2 n-1}\right] $
$ \times [C_{0}-C_{1} \frac{1}{x}+C_{2} \frac{1}{x^{2}}-\ldots . .+C_{2 n} \frac{1}{x^{2 n}}]$
Coefficient of $(\frac{1}{x})$ on the LHS
$ \begin{aligned} & =\text { Coefficient of } \frac{1}{x} \text { in } 2 n(\frac{1}{x^{2 n}})(1+x)^{2 n-1}(x-1)^{2 n} \\ & =\text { Coefficient of } x^{2 n-1} \text { in } 2 n\left(1-x^{2}\right)^{2 n-1}(1-x) \\ & =2 n(-1)^{n-1} \cdot(2 n-1) C_{n-1}(-1) \\ & =(-1)^{n}(2 n) \frac{(2 n-1) !}{(n-1) ! n !}=(-1)^{n} n \frac{(2 n) !}{(n !)^{2}} \cdot n \\ & \quad=-(-1)^{n} n \cdot C_{n} \quad …….(iii) \end{aligned} $
Again, the coefficient of $(\frac{1}{x})$ on the RHS
$ =-\left(C_{1}^{2}-2 \cdot C_{2}^{2}+3 \cdot C_{3}^{2}-\ldots-2 n C_{2 n}^{2}\right) $ $\quad$ …….(iv)
From Eqs. (iii) and (iv),
$ C_{1}^{2}-2 \cdot C_{2}^{2}+3 \cdot C_{3}^{2}-\ldots-2 n \cdot C_{2 n}^{2}=(-1)^{n} n \cdot C_{n} $
