Question

If $(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+\cdots +10(11{)}^9=k(10{)}^9$, then $k$ is equal to

• A. $\frac{121}{10}$
• B. $\frac{441}{100}$
• C. $100$
• D. $110$

Answer 1

The correct option is C $100$

Let the sum be denoted by S then

$S={10}^9+2\times 11\times {10}^8+3\times {11}^2\times {10}^7+.....+10\times {11}^9$

$or,\frac{11}{10}S=11\times {10}^8+2\times {11}^2\times {10}^7+.........+9\times {11}^9+11\times {11}^9$

on subtracting we get

$\frac{S}{10}={10}^9+11\times {10}^8+{11}^2\times {10}^7+......+{11}^8\times 10+{11}^9-{11}^{10}$

we can see that except the last term all the terms are in G.P with a =${10}^9$ and r=$\frac{11}{10}$

Hence $\Rightarrow \frac{S}{10}=\frac{{10}^9({\frac{11}{10}}^{10}-1)}{\frac{11}{10}-1}-{11}^{10}={10}^{10}({\frac{11}{10}}^{10}-1)-{11}^{10}\Rightarrow S={10\times 10}^{10}=100{\left(10\right)}^9$

Therefore, $k=100$