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 total number of ways of selecting two numbers from the set {1,2, 3, 4, ........3n} so that their sum is divisible by 3 is equal to a. (2n^2-n)/2 b. (3n^2-n)/2 c. 2n^2-n d. 3n^2-n

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 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

If a 1 , a 2 , a 3 , , a 2 n + 1 are in A.P., then a 2 n + 1 − a 1 a 2 n + 1 + a 1 + a 2 n − a 2 a 2 n + a 2 + + a n + 2 − a n a n + 2 + a n is equal to a. n ( n + 1 ) 2 × a 2 − a 1 a n + 1 b. n ( n + 1 ) 2 c. ( n + 1 ) ( a 2 − a 1 ) d. none of these

Let f(n)=1+1/2+1/3++1/ndot Then f(1)+f(2)+f(3)++f(n) is equal to (a) nf(n)-1 (b) (n+1)f(n)-n (c) (n+1)f(n)+n (d) nf(n)+n

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

If f(r)=1+1/2+1/3+1/ra n df(0)=0,t h e nsum_(r=1)^n(2r+1)f(r) is equal to (a) nf(n+1)-((n^2-3n+2))/2 (b) (n+1)^2f(n)-((n^2-3n+2))/2 (c) (n+1)^2f(n+1)-((n^2+3n+2))/2 (d) (n+1)^f(n+1)-((n^2-3n+2))/2

If there is a term independent of x in ( x+1/x^2)^n , show that it is equal to (n!)/((n/3)! ((2n)/3)! )

S_n=3/(1^2)+5/(1^2+2^2)+7/(1^2+2^2+3^2)+ upto n terms, then a. S_n=(6n)/(n+1) b. S_n=(6(n+2))/(n+1) c. S_oo=6 d. S_oo=1

The sum to n terms of the series 1+ 5 ((4n +1)/(4n - 3)) + 9 ((4n +1) /( 4n -3)) ^(2) + 13 ((4n +1)/( 4n -3)) ^(3) + ((4n + 1 )/( 4n -3)) ^(2)+ ……. is .