Quandaries
and Queries 

Prove that 1+3+5+...+(2n+1)= (n+1)^{2} for all n greater than or equal to 1. 

Hi Emma, Suppose that we use S to designate this sum, that is
There is a nice way to evaluate S that starts with evaluating 2S by writing the sum forwards and and then backwards.
Now add the terms in the sum by first adding down.
The terms in the sequence
are half the terms in the sequence
and thus the number of terms in the sequence
is ^{2n+2}/_{2} = n+1. Thus
and hence
Penny 

