Question
Fri March 09, 2012 By:

Link to the question :

Expert Reply
Fri March 09, 2012
Here Vn = { n + 3n-1 + 5n-2  + ........ n terms }
= n/2 [2n + (n-1)(2n-1)] = (2n3 - n2 + n)/2
 
Now V1 + V2 + ..... = sigma(Vn) = { 2sigma(n2) - sigma(n2) + sigma(n) }/2
= n(n+1)(3n2 +n+1)/12 = (b)
 
Also
Tn = n + (n-1)(2n-1) -2 = 2n(n-1) -1 which is always odd
And
Qn = Tn+1 , So we get Series of Q as 3, 11 , 23 , 39 ...... which is not in AP
Ask the Expert