Mathematics - Sequence & Series Question with Solution | TestHub

For (P), the sum of an infinite A.G.P. $a+(a+d)r+(a+2d)r^2+\dots$ is $\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$. Here $a=1,d=1,r=1/3$. So $S=\frac{1}{1-1/3}+\frac{1(1/3)}{(1-1/3{)}^2}=\frac{3}{2}+\frac{1/3}{4/9}=\frac{3}{2}+\frac{3}{4}=\frac{9}{4}$. So (P)-(1). For (Q), the series $1+2x+3x^2+\dots$ is the expansion of $(1-x{)}^{-2}=\frac{1}{(1-x{)}^2}$ for $|x|<1$. So (Q)-(2). For (R), the sum is ${\sum }_{k=1}^{10}k(k+1)={\sum }_{k=1}^{10}(k^2+k)=\frac{10(10+1)(10+2)}{3}=\frac{10\cdot 11\cdot 12}{3}=440$. So (R)-(3). For (S), the sum of cubes is ${\sum }_{k=1}^n{k}^3=[\frac{n(n+1)}{2}{]}^2$. So (S)-(4).