If the sum of first n terms of an A.P. is c n2, then the sum of squares of these n terms is

Question

If the sum of first n terms of an A.P. is c n2, then the sum of squares of these n terms is (A) (n(4 n2-1) c2/6) (B) (n(4 n2+1) c2/3) (C) (n(4 n2-1) c2/3) (D) (n(4 n2+1) c2/6)

Tardigrade Admin · Accepted Answer

Let Sn=c n2 Sn-1=c(n-1)2=c n2+c-2 c n ∴ Tn=2 c n-c (∵ Tn=Sn-Sn-1) Tn2=(2 c n-c)2=4 c2 n2+c2-4 c2 n ∴ Required sum, Sigma Tn2=(4 c2 ⋅ n(n+1)(2 n+1)/6)+n c2-2 c2 n(n+1) =(2 c2 n(n+1)(2 n+1)+3 n c2-6 c2 n(n+1)/3) =(n c2(4 n2+6 n+2+3-6 n-6)/3) =(n c2(4 n2-1)/3)

Q. If the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms is

$\frac{n ( 4 n ^{2} - 1 ) c ^{2}}{6}$

$\frac{n ( 4 n ^{2} + 1 ) c ^{2}}{3}$

$\frac{n ( 4 n ^{2} - 1 ) c ^{2}}{3}$

$\frac{n ( 4 n ^{2} + 1 ) c ^{2}}{6}$

Q. If the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is

6n(4n2−1)c2​

3n(4n2+1)c2​

3n(4n2−1)c2​

6n(4n2+1)c2​

Let Sn​=cn2 Sn−1​=c(n−1)2=cn2+c−2cn ∴Tn​=2cn−c(∵Tn​=Sn​−Sn−1​) Tn2​=(2cn−c)2=4c2n2+c2−4c2n ∴ Required sum, ΣTn2​=64c2⋅n(n+1)(2n+1)​+nc2−2c2n(n+1) =32c2n(n+1)(2n+1)+3nc2−6c2n(n+1)​ =3nc2(4n2+6n+2+3−6n−6)​ =3nc2(4n2−1)​

Q. If the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms is

$\frac{n ( 4 n ^{2} - 1 ) c ^{2}}{6}$

$\frac{n ( 4 n ^{2} + 1 ) c ^{2}}{3}$

$\frac{n ( 4 n ^{2} - 1 ) c ^{2}}{3}$

$\frac{n ( 4 n ^{2} + 1 ) c ^{2}}{6}$