As some even number problem's solution has already been provided so please find solution for following

First we'll verify the base case for $n = 1$ :

$\sum_{i=1}^1 i = \frac{1(1+1)}{2} = 1$

Now, we'll assume given statement is true up to $n - 1$.

So,

$\sum_{i=1}^{n-1} i = \frac{(n-1)(n-1+1)}{2} = \frac{(n-1)n}{2}$

To prove for the general case $n$,

$\sum_{i=1}^n i = \sum_{i=1}^{n-1} i + n = \frac{(n-1)n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}$

--Parth (talk) 05:07, 9 October 2015 (EDT)

First we'll verify the base case for $n = 1$ :

$\sum_{i=1}^1 i^3 = 1^3 = \frac{(1)^2(1+1)^2}{4} = 1$

Now, we'll assume given statement is true up to $n - 1$.
So, following statement is true.

$\sum_{i=1}^{n-1} i^3 = \frac{(n-1)^2(n-1+1)^2}{4} = \frac{(n-1)^2n^2}{4}$

To prove for the general case $n$,

\begin{align} \sum_{i=1}^{n} i^3 & = \sum_{i=1}^{n-1} i^3 + n^3 \\ & = \frac{(n-1)^2n^2}{4} + n^3 \\ & = \frac{(n^2 - 2n + 1)n^2 + 4n^3}{4} \\ & = \frac{n^4 - 2n^3 + n^2 + 4n^3}{4} \\ & = \frac{n^4 + 2n^3 + n^2}{4} \\ & = \frac{n^2(n^2 + 2n + 1)}{4} \\ \sum_{i=1}^{n} i^3 & = \frac{n^2(n + 1)^2}{4} \\ \end{align}

--Parth (talk) 05:07, 9 October 2015 (EDT)

First we'll verify the base case for $n = 1$ and we also assume $a=2$

$\sum_{i=0}^1 a^i = (2)^0 + (2)^1 = \frac{(2)^{1+1} - 1}{2 - 1} = 3$

Now, we'll assume given statement is true up to $n - 1$.
So, following statement is true.

$\sum_{i=0}^{n-1} a^i = \frac{a^{n-1+1}-1}{a-1} = \frac{a^n-1}{a-1}$

To prove for the general case $n$,

\begin{align} \sum_{i=0}^n a^i & = \sum_{i=0}^{n-1} a^i + a^n \\ & = \frac{a^n-1}{a-1} + a^n \\ & = \frac{a^n-1+a^n(a-1)}{a-1} \\ & = \frac{a^n-1+a^{n+1}-a^n}{a-1} \\ \end{align}

Thus,

$\sum_{i=0}^n a^i = \frac{a^{n+1}-1}{a-1}$

--Parth (talk) 05:07, 9 October 2015 (EDT)