Question

If $11+103+1005+\cdots n\text{terms}=\frac{{10}^{n+1}+x{n}^2+y}{9}$, then the value of $x-y$ is

• A. $-19$
• B. $19$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $19$

Let $S_n=11+103+1005+\cdots \text{to}n\text{terms}$
Now,
$\Rightarrow S_n=(10+1)+(100+3)+(1000+5)+\cdots \{{10}^n+(2n-1\left)\right\}\Rightarrow S_n=(10+{10}^2+{10}^3+\cdots +{10}^n)+(1+3+5+\cdots +(2n-1\left)\right)\Rightarrow S_n=[10\times \frac{({10}^n-1)}{10-1}+\frac{n}{2}(1+2n-1\left)\right]\Rightarrow S_n=\frac{10}{9}({10}^n-1)+n^2\Rightarrow S_n=\frac{{10}^{n+1}+9n^2-10}{9}\Rightarrow x=9,y=-10\therefore x-y=19$