Question

If $C_0+3C_1+5C_2+.....+(2n+1)C_n=32then{\int }_{-1}^1(n^2-n+1)dx=$

• A. $14$
• B. $3$
• C. $7$
• D. None of these

Answer 1

Answer: (A)

$14$

$C_0+3C_1+5C_2+....+(2n+1)C_n=32$
$\Rightarrow C_0+3C_1+5C_2+....+(2n+1)c_n=32$
$(2n+1)C_0+(2n-1)C_1+....+C_n=32$
By adding we get
$\Rightarrow (2n+1)\left[2^n\right]=32\times 2$
$\Rightarrow (n+1)2^n=32=(3+1)\left(2^3\right)$
$\Rightarrow n=3$
Now ${\int }_{-1}^1(n^2-n+1)dx$
$={\int }_{-1}^1(9-3+1)dx$
$=7{\int }_{-1}^{1}dx=7(x{)}_{-1}^1$
$=7(1+1)=14$