Let `A_n=(3/4)-(3/4)^2+(3/4)^3+….+(-1)^(n-1)(3/4)^n and B_n = 1-A_n`. find the least odd natural numbers `n_0`, so that `B_ngtA_n A` for all `ngen_0`

If quad a_(n)=(3)/(4)-((3)/(4))^(2)+((3)/(4))^(3)+...(-1)^(n-1)((3)/(4)) and b_(n)=1-a_(n), then find the minimum natural number n,such that b_(n)>a_(n)

Complete the following statements. and b_n=1-1_n then the smallest natural number n_0 , such that b_n > a_n !V n> n_0 is

1+5+9+...+(4n-3)=n(2n-1) for all natural numbers n

sum_(n=1)^(oo)((1)/(4n-3)-(1)/(4n-1))=(pi)/(n) find n

1^(3)+2^(3)+3^(3)+.....+n^(3)=(n(n+1)^(2))/(4), n in N

Let x_(n)=(2^(n)+3^(n))^(1//2n) for all natural number n. Then

If A={4^(n)-3n-1:n in N) and B={9(n-1):n in N} then

prove that 1+5+9+ . . .+(4n-3)=n(2n-1), for all natural number n.