Question

Find Sum upto n terms for the following series :
i) $1+2x+3x^2+4x^3+....,|x|<1$
ii) $1+4x+7x^2+10x^3+....,|x|<1$
iii) $2\times 1+3\times 2+4\times 4+5\times 8+....$

Answer 1

i) $1+2x+3x^2+4x^3+.......$

Let $S=1+2x+3x^2+4x^3+.......$

$x\cdot S=x+2x^2+3x^3+4x^4+......$

$S-x\cdot S=1+x+x^2+x^3+.......$

$S(1-x)=1+x+x^2+x^3+.......$

$S(1-x)=\frac{1}{1-x}$

$\therefore S=\frac{1}{(1-x{)}^2}$

ii) $1+4x+7x^2+10x^3+.......$

Let $S=1+4x+7x^2+10x^3+.......$

$x\cdot S=x+4x^2+7x^3+10x^4+......$

$S-x\cdot S=1+3x+3x^2+3x^3+.......$

$S-x\cdot S=1+3(x+x^2+x^3+.......)$

$(1-x)S=1+3\left(\frac{x}{1-x}\right)$

$S=\frac{1}{1-x}+\frac{3x}{(1-x{)}^2}$

$\therefore S=\frac{1+2x}{(1-x{)}^2}$

iii) $2\times 1+3\times 2+4\times 4+5\times 8+.........$

$2,3,4\to A.P\Rightarrow T_n=a+(n-1)d=2+(n-1)1=n+1$

$1,2,4\to G.P.\Rightarrow T_n=a{r}^{n-1}=1\cdot 2^{n-1}=2^{n-1}$

$\therefore T_n=(n+1)2^{n-1}$

$S=2\cdot 1+3\cdot 2+4\cdot 4+.............+T_n$

$\Rightarrow 2S=2\cdot +3\cdot 4+4\cdot 8+.........+T_{n-1}+T_n$

$S-2S=2+2+4+8+.......+T_n-T_{n-1}-T_n$

Again$2+4+8+.......+T_n-T_{n-1}$

$S=a\frac{r^{n-1}-1}{r-1}=2\frac{2^{n-1}-1}{2-1}=2\cdot 2^{n-1}-2$

$S=2^n-2$

$\therefore -S=2+2^n-2-T_n$

$\Rightarrow S=T_n-2^n$

substituting value of $T_n$ , we get

$S=(n+1)2^{n-1}-2^n$

$\Rightarrow S=2^{n-1}(n-1)$