JEE Main 2014 (Offline)Question 1
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 :
Sequence and Series · Mixed Progressions (AP-GP)
Solution
Formulas used. (1) For $S=\sum_{r=1}^{n}r\,x^{r-1}$: $S-xS=1+x+\cdots+x^{n-1}-n\,x^n$(2) $1+x+\cdots+x^{n-1}=\frac{x^n-1}{x-1}$Step 1 - Divide by $10^9$. With $x=\dfrac{11}{10}$: $$k=1+2x+3x^2+\cdots+10x^9 .$$ Step 2 - $k-xk$. $$(1-x)k=\left(1+x+\cdots+x^9\right)-10x^{10}=\frac{x^{10}-1}{x-1}-10x^{10}.$$ Step 3 - Put $x-1=\tfrac1{10}$. $$-\frac1{10}k=10\left(x^{10}-1\right)-10x^{10}=-10\ \Longrightarrow\ k=100 .$$ Answer: (A)