IIT JEE , AIEEE Forum - Mathematics , Algebra - Sum of 1 + (1+x) +(1+x+x^2)+(1+x+x^2+x^3)+(1+x+x^2+x^3+x^4).. up to n terms

sekara0809

Sum of 1 + (1+x) +(1+x+x^2)+(1+x+x^2+x^3)+(1+x+x^2+x^3+x^4).. up to n terms

IIT_JEE2008

x { [(xⁿ⁺¹ -1)/(x-1)] -[(n/(x-1)]}

Is tis d ans?

chinnu

on adding u get
n(1)+n-1(x)+n-2(x^2)+.....+1(x^n)
which is an agp and u can solve fm there.

allamraju

sriram.a

In simple way take S=n+(n-1)x+(n-2)x^2+.........................+x^n --------------------------------(1)

Sx=nx+(n-1)x^2+(n-2)x^3+.......................+x^(n+1)-------------------------------------(2)

subtract (2)-(1)

S(x-1)={x^(n+1)+x^n+............................+x+1}+(n-1)

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

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

allamraju

hsbhatt

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

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

$= x \ \left(\frac{x^n-1}{x-1} \right ) - n$

Hence

$S = x \ \left ( \frac{x^n-1}{(x-1)^2} \right) - \frac{n}{x-1}$