The sum to n terms of the series 1 + (1 + 3) + (1 + 3 +5 )+ ...... is ?

Please show the process also....!

1
thanks
but if we put n = 1 in your answer then it come -4 in place of 1... same in the case of 3,4,5 etc
yeah thats the reason i said that it is still wrong in previous comment
is the question entered correct?
yeah friend its 100% correct

Answers

2015-03-28T16:09:15+05:30
We will divide into 3 APs

(1).  1 (only single term) upto n terms...
=> sum = n

(2)      3 (only single term) upto  (n-1) terms
=> sum= ( \frac{n-1}{2} )*2(3)+(n-1-1)0               (d=0)
             = ( \frac{n-1}{2} )6

(3)    5 (only one term) upto (n-2 ) terms
=>sum =  ( \frac{n-2}{2} )*2(5)                           (d=0)
             = ( \frac{n-2}{2} )*10

summing up all the APs

=> n+ (n-1)/2*6 +(n-2)/2*10
=>n+ 3(n-1)+5(n-2)
=>n+3n-3+5n-10
=.9n-13

0
Please let me know the answer
answer is n^2
I have solved it....!:)
Let the last term of each term of above A.P be l
then l = a +(n-1)d => here a = 1 and d = 2
therefore, l= 2n-1

Now the sum of above A.P will be = n/2(a + l)
where a= 1, l = 2n-1
So, sum of n terems = n/2[1 + (2n -1)] = n(n) = n^2