We have a sequence of numbers xn determined by the equality

xn=xn−1+xn−22

The first and zeroth term are x1 and x0.The following limit must be expressed in terms of x0 and x1

limn→∞xnThe options are:

A)x0+2x13

B)2x0+2x13C)2x0+3x13

D)2x0−3x13Since it was a multiple choice exam I plugged x0=1 and x1=1. Which means that all terms of this sequence is 1,i.e, xn=1,n∈N

From this I concluded that option A was correct.I could not find any way to solve this one hence I resorted to this trick. What is the actual method to find the sequence’s limit?

**Answer**

2xn=xn−1+xn−2

2x2=x1+x02x3=x2+x12x4=x3+x22x5=x4+x3...2xn=xn−1+xn−2

Now sum every equation and get

2xn+xn−1=2x1+x0

Supposing that xn has a limit L then making n→∞ we get:

2L+L=2x1+x0→L=2x1+x03

**Attribution***Source : Link , Question Author : Ananth Kamath , Answer Author : Arnaldo*