[" (A) "6quad " (B) "2],[" 14."2+4+7+11+16+........." to "n" terms "=T_(n)=(n(n+1))/(2)+1],[" (C) "(1)/((n^(2)-3n+8))" (D) "(n)/(6)(n^(2)-3n+1)]

Prove that log_(e)((n^(2))/(n^(2)-1))=(1)/(n^(2))+(1)/(2n^(4))+(1)/(3n^(6))+

(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+... n terms =(A)(n)/(6n+4) (B) (n)/(3n+2)(C)(n)/(4n+6)(D)(1)/(2(2n+3))

1.3+3.5+5.7+......+(2n-1)(2n+1)=(n(4n^(2)+6n-1))/(3)

1+(n)/(2)+(n(n-1))/(2.4)+(n(n-1) (n-2))/(2.4.6)+…....=

1.3+2.4+3.5+....+n(n+2)=(n(n+1)(2n+7))/(6)

If the sum of first n terms of an AP is cn^(2) , then the sum of squares of these n terms is (2009)( a) (n(4n^(2)-1)c^(2))/(6) (b) (n(4n^(2)+1)c^(2))/(3) (c) (n(4n^(2)-1)c^(2))/(3) (d) (n(4n^(2)+1)c^(2))/(6)

1.3 + 2.4 + 3.5 + …… + n(n+2) =( n(n+1)(2n+7))/6

Find the sum of n terms of the series (1)/(2*4)+(1)/(4*6)+... (A) (n)/(n+1) (B) (n)/(4(n+1)) (C) (1)/((2n)(2n+2))( D) )(1)/(2^(n)(2^(n)+2))

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 (2009) (n(4n^2-1)c^2)/6 (b) (n(4n^2+1)c^2)/3 (n(4n^2-1)c^2)/3 (d) (n(4n^2+1)c^2)/6

Simplify (6(8)^(n+1)+16(2)^(3n-2))/(10(2)^(3n+1)-7(8)^(n))