If `((2+1)(2^2+1)(2^4+1)(2^8+1))/(2^8-1)=4^n+1`, then n is:

Find the value of (-1)^(n)+(-1)^(2n)+(-1)^(2n_1)+(-1)^(4n+2) , where n is any positive and integer.

2^32-(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1) is equal to- 2^32-(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1) के बराबर है।

[ If 2 is the sum of infinity of a G.P.,whose first clement is 1 ,then the sum of the first n terms is [ 1) (2^(n)-1)/(2^(n)), 2) (2^(n)-1)/(2^(n-1)), 3) (2^(n-1)-2)/(2), 4) (2^(n-1)-1)/(2^(n))]]

If the sum of the series 1+(3)/(2)+(5)/(4)+(7)/(8)+……+((2n-1))/((2)^(n-1)) is f(n) , then the value of f(8) is

1 4/(8)-:1(2)/(8)xx1(2)/(4)+2(1)/(2)-1(1)/(4)

Find lim_ (n rarr oo) [(n ^ (4) +1) ^ ((1) / (2)) - (n ^ (4) -1) ^ ((1) / (2))] -: n ^ (- 2)

If (4^(n+1).2^n-8^n)/(2^(3m))=3/8 then prove that n+1=m

The sum of the series: (1)/(log_(2)4)+(1)/(log_(4)4)+(1)/(log_(8)4)+...+(1)/(log_(2n)4) is (n(n+1))/(2) (b) (n(n+1)(2n+1))/(12) (c) (n(n+1))/(4) (d) none of these