Question

${\left({}^{2n}{c}_0\right)}^2-{\left({}^{2n}{c}_1\right)}^2+{\left({}^{2n}{c}_2\right)}^2-.....+{(-1)}^{2n}{\left({}^{2n}{c}_{2n}\right)}^2$ equal to

• A. ${\left({}^{2n}{c}_n\right)}^2$
• B. ${(-1)}^{2n}\left({}^{2n}{c}_n\right)$
• C. ${(-1)}^n\left({}^{2n}{c}_n\right)$
• D. $\left({}^n{c}_n\right)+\left({}^{2n}{c}_n\right)$

Answer 1

The correct option is B ${(-1)}^{2n}\left({}^{2n}{c}_n\right)$

$\begin{array}{cc}(1+x{)}^{2n}=& (2n{)}_{C_0}(x{)}^{2n}+(2n\left)c_1\right(x{)}^{2n-1}+\cdots \cdots +(2n)C_{2n}\cdots (1)\end{array}$

$(1-x{)}^{2n}=(2n{)}_{C_0}-\left(2n{)}_{C_1}\right(x{)}^2+\left(2n{)}_{C_2}\right(x{)}^2+\cdots +\left(2n\right)C_{2n}{\left(x\right)}^{2n}\cdots \left(2\right)$

Multiplying equation 1 and 2

$\begin{array}{cc}(1+x{)}^{2n}\cdot (1-x{)}^{2n}& =\left\{\right(2n\left)c_0\right(x{)}^{2n}+\left(2n\right)c_1(x{)}^{2n-1}+\cdots \cdots +(2n\left)c_{2n}\right\}\\ & \times \left\{\right(2n)c_0-(2n\left)c_1\right(x{)}^2+\dots \dots +\left(2n\right)c_{2n}\left(x{)}^{2n}\right\}\end{array}$

$\begin{array}{cc}\Rightarrow \left[\right(1+x\left)\right(1-x){]}^{2n}& ={(1-x^2)}^{2n}\\ \text{Now}{(1-x^2)}^{2n}& =\sum _{r=0}^{2n}\left(2n\right)c_r{(-x^2)}^r\\ & =\sum _{r=0}^{2n}(-1{)}^r(2n\left)c_r\right(x{)}^{2r}\end{array}$

$\begin{array}{cc}\sum _{r=0}^{2n}(-1{)}^r(2n\left)c_r\right(x{)}^{2r}& =\left\{\right(2n\left)c_0\right(x{)}^{2n}+\left(2n\right)c_L(x{)}^{2n-1}+\cdots +(2n)c_n\\ & \times \left\{\right(2n)c_0-(2n)c_{1}x+\cdots +(2n\left)c_{2n}\right(x{)}^{2n}\}\end{array}$

Now for coefficient of $x^{2n}$

putting r = n in LHS of above equation

$\begin{array}{cc}(-1{)}^n(2n)c_n=& {\left\{\right(2n\left)c_0\right\}}^2-{\left\{\right(2n\left)c_1\right\}}^2+{\left\{\right(2n\left)c_2\right\}}^2+\dots \dots \dots (-1{)}^n{\left\{\right(2n\left)c_{2n}\right\}}^2\end{array}$

So answer option (B).