# How to prove this mathematical induction?

Can someone explain this prove mathematical induction question?

Use mathematical induction to prove the following expression: $$\sum_{k = 1}^{n} 2^{n + 1} - 1.$$ I tried my best to solve it but when i tried to prove for p(1) it got failed and for p(0) it is working fine.

I am new to mathematical induction and have test tomorrow.

I am new to this maths stack exchange and please for give if there is any formatting mistakes.

Tried to solve this

Tried to solve this - Continued

Please help me out and thanks.

• Note that $\sum_{i=1}^n a_i= a_n+\sum_{i=1}^{n-1} a_i$ for $n\ge1$ by the only way to rigorously define $\sum$ – Hagen von Eitzen Nov 11 '18 at 13:52
• Maybe your problem is related to the fact that the given identity is wrong. Try with the correct version that is $\sum_{i=0}^n2^i=2^{n+1}-1$. – gimusi Nov 11 '18 at 13:55

## 1 Answer

There is a typo in the text, it should be

$$\sum_{i=0}^n2^i=2^{n+1}-1$$

then proceed firstly by base case and then by induction step assuming true $$P(n)$$ and deriving from it $$P(n+1)$$.

Refer also to the related

• or it should be $\ldots =2^{n+1}-2$ – Hagen von Eitzen Nov 11 '18 at 13:52
• @HagenvonEitzen Equivalentelly yes of course, but referring to the usual expression for the geometric series I think that the typo is likely to be that for $i=0$. – gimusi Nov 11 '18 at 13:54
• @gimusi Thanks for the answering the query. Can you please also clear the solution once ? I know i should do it myself but it has become very confusing now. Can you ? – Himanshu Chawla Nov 11 '18 at 13:59
• @HimanshuChawla No I can't but I can give you a further tip. Start from $$\sum_{i=0}^{n+1}2^i=2^{n+1}+\sum_{i=0}^n2^i=\ldots$$ and try to show that $$\sum_{i=0}^{n+1}2^i=2^{n+2}-1$$ Also take a look to others induction proof you can find here on MSE and practice a lot on that. – gimusi Nov 11 '18 at 14:03
• @HimanshuChawla Refer for example to OP – gimusi Nov 11 '18 at 14:08