(1) + (1 + 3) + (1 + 3 + 5) + upto ‘n’ terms = ………A) \(\cfr

1.	(1) + (1 + 3) + (1 + 3 + 5) + upto ‘n’ terms = ………A) \(\cfrac{n(n-1)(2n-1)}{24}\)B) \(\cfrac{n(n+1)(2n-1)}{12}\)C) \(\cfrac{n(n+1)(2n+1)}{6}\)D) \(\cfrac{n(n+1)(2n+1)}{24}\)
Answer» Correct option is (C) \(\frac{n(n+1)(2n+1)}{6}\) (1) + (1+3) + (1+3+5) + ....... + upto n terms = 1 + (1+3) + (1+3+5) + ....... + (1+3+5+.....+n terms) = 1 + (1+3) + (1+3+5) + ....... + (1+3+5+.....+ 2n - 1) = 1 + (1+3) + (1+3+5) + ....... + \(\frac n2(1+(2n-1))\) \((\because S_n=\frac n2(a+a_n)\) in A.P.) = 1 + (1+3) + (1+3+5) + ....... + \(n^2\) \(=\sum n^2\) \(=\frac{n(n+1)(2n+1)}{6}\) Correct option is C) \(\cfrac{n(n+1)(2n+1)}{6}\)

(1) + (1 + 3) + (1 + 3 + 5) + upto ‘n’ terms = ………A) \(\cfrac{n(n-1)(2n-1)}{24}\)B) \(\cfrac{n(n+1)(2n-1)}{12}\)C) \(\cfrac{n(n+1)(2n+1)}{6}\)D) \(\cfrac{n(n+1)(2n+1)}{24}\)

Answer»

Correct option is (C) \(\frac{n(n+1)(2n+1)}{6}\)

(1) + (1+3) + (1+3+5) + ....... + upto n terms

= 1 + (1+3) + (1+3+5) + ....... + (1+3+5+.....+n terms)

= 1 + (1+3) + (1+3+5) + ....... + (1+3+5+.....+ 2n - 1)

= 1 + (1+3) + (1+3+5) + ....... + \(\frac n2(1+(2n-1))\)

\((\because S_n=\frac n2(a+a_n)\) in A.P.)

= 1 + (1+3) + (1+3+5) + ....... + \(n^2\)

\(=\sum n^2\)

\(=\frac{n(n+1)(2n+1)}{6}\)

Correct option is C) \(\cfrac{n(n+1)(2n+1)}{6}\)