[sum n=10,Sigma n^(3)=..........],[[" A) "100," B) "1001" C) "200," D) "80]]

sum n = 10, sum n^(3) = ……………

sum n = 5 . Then sum n^3 =

If Sigma n = 210 " then " Sigma n^(2) =…….

For all n in N, Sigma n

For all n in N, Sigma n

.The sum to 'n' terms of the series 9 + 99 + 999 ... ... is ] , [ 1) (10 ( 10^ n − 1 ) − n) 2) (10/ 9 [ 10^ n − 1 ] + n) 3) (10/ 9 [ 10^ n − 1 ] − n) 4) (10/ 9 [ 10^ n − 1) ]

The sum of series Sigma_(n=1)^(oo) (2n)/(2n+1)! is

sum_ (n = 1) ^ (oo) (5 ^ (n) -2 ^ (n)) / (10 ^ (n))

"Evaluate "Sigma_(n=1)^(6) (3+2^(n)).