JEE Main 2004Annual, Paper 1MathematicsMediumSequence and Series179× since 2002The sum of the first n terms of the series 12+2.22+32+2.42+52+2.62+.... is n(n+1)22{1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + ....\,is\,{{n{{(n + 1)}^2}} \over 2}12+2.22+32+2.42+52+2.62+....is2n(n+1)2 when n is even. When n is odd the sum isAOption A. [n(n+1)2]2{\left[ {{{n(n + 1)} \over 2}} \right]^2}[2n(n+1)]2BOption B. n2(n+1)2{{{n^2}(n + 1)} \over 2}2n2(n+1)COption C. n(n+1)24{{n{{(n + 1)}^2}} \over 4}4n(n+1)2DOption D. 3n(n+1)2\,{{3n(n + 1)} \over 2}23n(n+1)CheckSkip · show solution
