# If #sum_(n=2) ^oo (1+k)^-n=2# what is #k#?

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**The answer is supposed to be **#k=(sqrt3-1)/2# I just don't know how to get there.

**The answer is supposed to be #k=(sqrt3-1)/2# I just don't know how to get there.**

##### 1 Answer

Dec 14, 2016

#### Explanation:

A geometric series of the form

With that,

#=-(1/(1+k))^0-(1/(1+k))^1+sum_(n=0)^oo(1/(1+k))^n#

#=-1-1/(1+k)+1/(1-(1/(1+k)))#

#=-1-1/(1+k)+1/(k/(1+k))#

#=-1-1/(1+k)+(1+k)/k#

#=2#

We can now solve for

But we must have