If `1+(1+2)/2+(1+2+3)/3+ddotto\ n` terms is `Sdot` Then, `S` is equal to a. `(n(n+3))/4` b. `(n(n+2))/4` c. `(n(n+1)(n+2))/6` d. `n^2`

The sum of the series 1^2+3^2+5^2+ddotto\ n terms is a. (n(n+1)(2n+1))/2 b. (n(2n-1)(2n+1))/3 c. ((n-1)^2(2n+1))/6 d. ((2n+1)^3)/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

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

If n is a non zero rational number then show that 1 + n/2 + (n (n - 1))/(2.4) + (n(n-1)(n - 2))/(2.4.6) + ….. = 1 + n/3 + (n (n + 1))/(3.6) + (n (n + 1) (n + 2))/(3.6.9) + ….

If n is a non zero rational number then show that 1 + n/2 + (n (n - 1))/(2.4) + (n(n-1)(n - 2))/(2.4.6) + ….. = 1 + n/3 + (n (n + 1))/(3.6) + (n (n + 1) (n + 2))/(3.6.9) + ….

1.2.3+2.3.4++n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/(4)

1.2.3+2.3.4+....+n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/4

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

The sum of the series 1+4+3+6+5+8+ upto n term when n is an even number (n^2+n)/4 2. (n^2+3n)/2 3. (n^2+1)/4 4. (n(n-1))/4 (n^2+3n)/4