Question

If $I_n=\int ta{n}^{n}xdx,then{I}_0+I_1+2(I_2+...I_8)+I_9+I_{10}$, is equal to

• A. ${\sum }_{n=1}^9\frac{ta{n}^{n}x}{n}$
• B. $1+{\sum }_{n=1}^8\frac{ta{n}^{n}x}{n}$
• C. ${\sum }_{n=1}^9\frac{ta{n}^{n}x}{n+1}$
• D. ${\sum }_{n=2}^{10}\frac{ta{n}^{n}x}{n+1}$

Answer 1

The correct option is A ${\sum }_{n=1}^9\frac{ta{n}^{n}x}{n}$

We have $I_n=\int ta{n}^{n}xdx=\int ta{n}^{n-2}x(se{c}^{2}x-1)dx=\frac{ta{n}^{n-1}x}{n-1}-I_{n-2}i.e.I_{n-2}+I_n=\frac{ta{n}^{n-1}x}{n-1}(n\ge 2)$
Thus, we have
$I_0+I_1+2(I_2+....+I_8)+I_9+I_{10}=(I_0+I_2)+(I_1+I_3)+(I_2+I_4)+(I_3+I_5)+(I_4+I_6)+(I_5+I_7)+(I_6+I_8)+(I_7+I_9)+(I_8+I_{10}).={\sum }_{n=2}^{10}\frac{ta{n}^{n-1}x}{n-1}={\sum }_{n=1}^9\frac{ta{n}^{n}x}{n}$