`1, 5, 9, 13...` Use mathematical induction to find a formula for the sum of the first n terms of the sequence.

`1,5,9,13,......`

Now let's denote the first term` ` of the above sequence by `a_1` and k'th term by `a_k` and let's write down the first few sums of the sequence.

`S_1=1`

`S_2=1+5=6=(2(1+5))/2=(2(a_1+a_2))/2`

`S_3=1+5+9=15=(3(1+9))/2=(3(a_1+a_3))/2`

`S_4=1+5+9+13=28=(4(1+13))/2=(4(a_1+a_4))/2`

From the above sequence it appears that the formula for the sum of the k terms is,

`S_k=(k(1+a_k))/2`

Now the difference between the consecutive terms of the series is 4, so we can write down ,

`a_k=1+(k-1)4=1+4k-4=4k-3`

`a_(k+1)=4(k+1)-3=4k+4-3=4k+1`

`S_k=(k(1+4k-3))/2=(k(4k-2))/2`

`S_k=k(2k-1)`

Now...

`1,5,9,13,......`

Now let's denote the first term` ` of the above sequence by `a_1` and k'th term by `a_k` and let's write down the first few sums of the sequence.

`S_1=1`

`S_2=1+5=6=(2(1+5))/2=(2(a_1+a_2))/2`

`S_3=1+5+9=15=(3(1+9))/2=(3(a_1+a_3))/2`

`S_4=1+5+9+13=28=(4(1+13))/2=(4(a_1+a_4))/2`

From the above sequence it appears that the formula for the sum of the k terms is,

`S_k=(k(1+a_k))/2`

Now the difference between the consecutive terms of the series is 4, so we can write down ,

`a_k=1+(k-1)4=1+4k-4=4k-3`

`a_(k+1)=4(k+1)-3=4k+4-3=4k+1`

`S_k=(k(1+4k-3))/2=(k(4k-2))/2`

`S_k=k(2k-1)`

Now we can verify that it is valid for n=1,

Plug in k=1 to verify,

`S_1=1(2*1-1)=1`

Let's assume that the formula is valid for n=k and now we have to show that it is valid for n=k+1

`S_(k+1)=1+5+9+13........+a_k+a_(k+1)`

`S_(k+1)=S_k+a_(k+1)`

`S_(k+1)=k(2k-1)+4k+1`

`S_(k+1)=2k^2-k+4k+1`

`S_(k+1)=2k^2+3k+1`

`S_(k+1)=2k^2+2k+k+1`

`S_(k+1)=2k(k+1)+1(k+1)`

`S_(k+1)=(k+1)(2k+1)`

`S_(k+1)=(k+1)[2(k+1)-1]`

So the formula is valid for n=k+1 also, 

So the formula can be written as ,

`S_n=n(2n-1)`