NIMCET 2017 Question Paper with solutions

We have,
${\left(1-\mathit{x}+{\mathit{x}}^2\right)}^{\mathit{n}}$ = ${\mathit{a}}_0+{\mathit{a}}_1\mathit{x}+{\mathit{a}}_2{\mathit{x}}^2$ + ... + ${\mathit{a}}_{2\mathit{n}}{\mathit{x}}^{2\mathit{n}}$ ... (i)
Put x = 1 in Eq. (i), we get
${\left(1-1+1\right)}^{\mathit{n}}$ = ${\mathit{a}}_0+{\mathit{a}}_1+{\mathit{a}}_2$ + ... + ${\mathit{a}}_{2\mathit{n}}$
⇒ ${\mathit{a}}_0+{\mathit{a}}_1+{\mathit{a}}_2$ ... ${\mathit{a}}_{2\mathit{n}}$ = 1 ... (ii)
Again, put x = - 1 in Eq. (i), we get
${\left(1-\left(-1\right)+{\left(-1\right)}^2\right)}^{\mathit{n}}$ = ${\mathit{a}}_0-{\mathit{a}}_1+{\mathit{a}}_2$ - ... + ${\mathit{a}}_{2\mathit{n}}$
⇒ ${\mathit{a}}_0-{\mathit{a}}_1+{\mathit{a}}_2$ - ... + ${\mathit{a}}_{2\mathit{n}}$ = $3^{\mathit{n}}$ ... (iii)
On adding Eq. (ii) and Eq. (iii), we get
$2\left({\mathit{a}}_0+{\mathit{a}}_2+...+{\mathit{a}}_{2\mathit{n}}\right)$ = $3^{\mathit{n}}$ + 1
⇒ ${\mathit{a}}_0+{\mathit{a}}_2$ + ... + ${\mathit{a}}_{2\mathit{n}}$ = $\frac{3^{\mathit{n}}+1}{2}$