Progression (Sequence and Series)Hard
Question
If the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is
Options
A.
B.
C.
D.
Solution
tn = c {n2 - (n - 1)2}
= c (2n - 1)
⇒ tn2 = c2 (4n2 - 4n + 1)
⇒
=
{4(n + 1)(2n + 1)- 12(n + 1) + 6}
=
{4n2 + 6n + 2 - 6n - 6 + 3} = 
n(4n2 - 1)
= c (2n - 1)
⇒ tn2 = c2 (4n2 - 4n + 1)
⇒
=
{4(n + 1)(2n + 1)- 12(n + 1) + 6}=
{4n2 + 6n + 2 - 6n - 6 + 3} = n(4n2 - 1)Create a free account to view solution
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