`sum_(k=1)^(6)((2+4+6+...+2k)^(2))/(1+3+5+...+(2k-1))` divided by 13,then the remainder is

sum_ (k = 1) ^ (n) (2 ^ (k) + 3 ^ (k-1))

The remainder when (sum_(k=1)^(5) ""^(20)C_(2k-1))^(6) is divided by 11, is :

sum_(k=1)^(3n)6nC_(2k-1)(-3)^(k) is equal to

If sum_(k=1)^(k=n)(tan^(-1)(2k))/(2+k^(2)+k^(4))=tan^(-1)((6)/(7)), then the value of ' 'n' is equal to

Find sum_(k=1)^(n)(tan^(-1)(2k))/(2+k^(2)+k^(4))

sum_ (k = 1) ^ (6) (sin, (2 pi k) / (7) -i cos, (2 pi k) / (7)) =?

If the value of the sum n^(2) + n - sum_(k = 1)^(n) (2k^(3)+ 8k^(2) + 6k - 1)/(k^(2) + 4k + 3) as n tends to infinity can be expressed in the form (p)/(q) find the least value of (p + q) where p, q in N

sum_(k=1)^(20)k[(1)/(k)+(1)/(k+1)+(1)/(k+2)+......+(1)/(20)]