23simply put n =1
so first termof AP = c
since sum of square of 1st term of Ap =c2
and only option c gives ans as c2 hence option C shud be the an s
If the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is
A) \frac{n(4n^{2}-1)c^{2}}{6} B) \frac{n(4n^{2}+1)c^{2}}{3} C) \frac{n(4n^{2}-1)c^{2}}{3} D) \frac{n(4n^{2}+1)c^{2}}{6}
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3 Answers
62The fact is simple.. without solving..
SUm of first n odd numbers is n^2 (This is a very old observation)
WHat we do is take c, 3c, 5c, ... and so on
what we have to do is to find
(c)^2 {12+32+52+....+(2n-1)2}
Can you first prove the fact that I said (not a veyr tough thing...)
Second find the sum that I have given
1another way
S(n)=cn^2
T(n)=c(2n-1)
hence we got what nishant sir told
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