Question

If $a_0,a_1,a_2,......$ be the coefficients in the expansion of $(1+x+x^2{)}^n$ is ascending powers of $x$, then $a_0^2-a_1^2+a_2^2-a_3^2+......+a_{2a}^2=a_n$.

• A. True
• B. False

Answer 1

The correct option is A True

Given $a_0$, $a_1$, $a_2$...........$a_{2n}$ are coefficients of x in ascending order.

Thus, we can write,

${(1+x+x^2)}^n=a_0+a_{1}x+a_2{x}^2+.........+a_n{x}^n+..............+a_{2n}{x}^{2n}$ (1)

1) Put $x=-1$ in equation (1), we get,

${(1+(-1)+{(-1)}^2)}^n=a_0+a_1(-1)+a_2{(-1)}^2+.........+a_{2n}{(-1)}^{2n}$

${(0+1)}^n=a_0-a_1+a_2-a_3+.........+a_{2n}$

$\therefore 1=a_0-a_1+a_2-a_3+.........+a_{2n}$ (2)

2) Put $n=\frac{-1}{x}$ in equation (1), we get,

${(1+\left(\frac{-1}{x}\right)+{\left(\frac{-1}{x}\right)}^2)}^n=a_0+a_1\left(\frac{-1}{x}\right)+a_2{\left(\frac{-1}{x}\right)}^2+.........+a_{2n}{\left(\frac{-1}{x}\right)}^{2n}$

$\therefore {(1-\frac{1}{x}+\frac{1}{x^2})}^n=a_0-\frac{a_1}{x}+\frac{a_2}{x^2}+.........+\frac{a_{2n}}{x^{2n}}$

Multiplying both sides by ${(1+x+x^2)}^n$, we get,

${(1+x+x^2)}^n{(1-\frac{1}{x}+\frac{1}{x^2})}^n={(1+x+x^2)}^n[a_0-\frac{a_1}{x}+\frac{a_2}{x^2}+.........+\frac{a_{2n}}{x^{2n}}]$

$\therefore {(1+x+x^2)}^n{(1-\frac{1}{x}+\frac{1}{x^2})}^n=[a_0+a_{1}x+a_2{x}^2+......+a_{2n}{x}^{2n}][a_0-\frac{a_1}{x}+\frac{a_2}{x^2}+.........+\frac{a_{2n}}{x^{2n}}]$

$\therefore {\left[(1+x+x^2)(1-\frac{1}{x}+\frac{1}{x^2})\right]}^n={a_0}^2-{a_1}^2+{a_2}^2-{a_3}^2+..........+{a_{2n}}^2$

$\therefore {\left[(1+x+x^2)\left(\frac{x^2-x+1}{x^2}\right)\right]}^n={a_0}^2-{a_1}^2+{a_2}^2-{a_3}^2+..........+{a_{2n}}^2$

$\therefore {\left[\frac{((1+x^2)+x)((1+x^2)-x)}{x^2}\right]}^n={a_0}^2-{a_1}^2+{a_2}^2-{a_3}^2+..........+{a_{2n}}^2$

$\therefore {\left[\frac{{(1+x^2)}^2-x^2}{x^2}\right]}^n={a_0}^2-{a_1}^2+{a_2}^2-{a_3}^2+..........+{a_{2n}}^2$

$\therefore {\left[\frac{1+2x^2+x^4-x^2}{x^2}\right]}^n={a_0}^2-{a_1}^2+{a_2}^2-{a_3}^2+..........+{a_{2n}}^2$

$\therefore {\left[\frac{1+x^2+x^4}{x^2}\right]}^n={a_0}^2-{a_1}^2+{a_2}^2-{a_3}^2+..........+{a_{2n}}^2$ (3)

Consider Left Hand Side of equation (3) i.e. ${\left[\frac{1+x^2+x^4}{x^2}\right]}^n$

Put $x^2=y$ in above equation, we get,

$LHS={\left(\frac{1+y+y^2}{y}\right)}^n$

$\therefore LHS=\frac{{(1+y+y^2)}^n}{y^n}$

From equation (1), we can write,

$\therefore LHS=\frac{a_0+a_{1}y+a_2{y}^2+.........+a_n{y}^n+..............+a_{2n}{y}^{2n}}{y^n}$

$\therefore LHS=\frac{a_0}{y^n}+\frac{a_{1}y}{y^n}+\frac{a_2{y}^2}{y^n}+.............+\frac{a_n{y}^n}{y^n}+..................\frac{a_{2n}{y}^{2n}}{y^n}$

$\therefore LHS=\frac{a_0}{y^n}+\frac{a_{1}y}{y^n}+\frac{a_2{y}^2}{y^n}+.............+a_n+..................\frac{a_{2n}{y}^{2n}}{y^n}$

As entire RHS is constant, we have to consider constant term on LHS to equate with RHS.

From above equation, it is clear that only constant term in above equation is $a_n$

Thus, from equation (3),

${a_0}^2-{a_1}^2+{a_2}^2-{a_3}^2+..........+{a_{2n}}^2=a_n$