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 $2 n(1+x)^{2 n-1}=C _1+2 \cdot C _2 x+3 \cdot C _3 x^{2}$
$$ +\ldots+2 n C _{2 n} x^{2 n-1} $$
and $1-\frac{1}{x}^{2 n}=C _0-C _1 \cdot \frac{1}{x}+C _2 \cdot \frac{1}{x^{2}}-C _3 \cdot \frac{1}{x^{3}}$
$$ +\ldots+C _{2 n} \cdot \frac{1}{x^{2 n}} $$
On multiplying Eqs. (i) and (ii), we get
$$ \begin{array}{rl} 2 n(1+x)^{2 n-1} & 1-\frac{1}{x}^{2 n} \\ =\left[C _1+2\right. & \left.\cdot C _2 x+3 \cdot C _3 x^{2}+\ldots+2 n \cdot C _{2 n} x^{2 n-1}\right] \\ & \times C _0-C _1 \frac{1}{x}+C _2 \frac{1}{x^{2}}-\ldots . .+C _{2 n} \frac{1}{x^{2 n}} \end{array} $$
Coefficient of $\frac{1}{x}$ on the LHS
$$ \begin{aligned} & =\text { Coefficient of } \frac{1}{x} \text { in } 2 n \frac{1}{x^{2 n}}(1+x)^{2 n-1}(x-1)^{2 n} \\ & =\text { Coefficient of } x^{2 n-1} \text { in } 2 n\left(1-x^{2}\right)^{2 n-1}(1-x) \\ & =2 n(-1)^{n-1} \cdot(2 n-1) C _{n-1}(-1) \\ & =(-1)^{n}(2 n) \frac{(2 n-1) !}{(n-1) ! n !}=(-1)^{n} n \frac{(2 n) !}{(n !)^{2}} \cdot n \\ & \quad=-(-1)^{n} n \cdot C _n \end{aligned} $$
Again, the coefficient of $\frac{1}{x}$ on the RHS
$$ =-\left(C _1^{2}-2 \cdot C _2^{2}+3 \cdot C _3^{2}-\ldots-2 n C _{2 n}^{2}\right) $$
From Eqs. (iii) and (iv),
$$ C _1^{2}-2 \cdot C _2^{2}+3 \cdot C _3^{2}-\ldots-2 n \cdot C _{2 n}^{2}=(-1)^{n} n \cdot C _n $$
