Question

find the value of c if sigma n = 2 to infinity of (1+c)^-n = 2

Answer 1

simplify first with
\[y=\frac{1}{1+c}=(1+c)^{-1}\]
\[\sum_{n=2}^\infty (1+c)^{-n}=2\\\sum_{n=2}^\infty y^n=2\]
\[\large \frac{1}{1-y}=\sum_{n=0}^\infty y^n=1+y+(y^2+y^3+\cdots)=1+y+\sum_{n=2}^\infty y^n\]

Answer 2

\[\large \sum_{n=2}^\infty y^n=\frac{1}{1-y}-1-y\]
\[\large {2=\sum_{n=2}^\infty y^n=\frac{1}{1-y}-1-y\\2=\frac{1}{1-y}-1-y}\] solve y, then solve c