How do I calculate this sum in terms of 'n'? I know this is a harmonic progression, but I can't find how to calculate the summation of it. Also, is it an expansion of any mathematical function? 1 ...
What is interesting is that your formula is the closed form for a different summation, i.e. $\displaystyle \sum_ {i=0}^n \binom {i+1}2=\sum_ {i=0}^n \frac {i (i+1)}2=\frac {n (n+1) (n+2)}6=\binom {n+2}3$.
How do I prove this by induction? Prove that for every natural number n, $ 2^0 + 2^1 + ... + 2^n = 2^{n+1}-1$ Here is my attempt. Base Case: let $ n = 0$ Then, $2^{0+1} - 1 = 1$ Which is true.
The first chapter of Concrete Mathematics by Graham, Knuth, and Patashnik presents about seven different techniques for deriving this identity, so you might be interested to look at that.
I know that the sum of powers of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place. For example, sum of n numbers is ...
