Precalculus, Chapter 9, 9.4, Section 9.4, Problem 41

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)