ত্রিভুজের 20তম সারিতে\n1\n 2 3 \n\n4\n5\n6\n7\n10\n(A) শেষ পদ = 210\n(B) প্রথম পদ...

n/2 - (3n)/4 +(5n)/6 = 21

If n/2 - (3n)/4 + (5n)/6 = 21 , then n = ?

If n(AuuBuuC) + n(AuuBuuC)' = 50, n(A)=10, n(B)=15, n(C)=20,n(AnnB)=5,n(BnnC) =6, n(CnnA)=10 and n(AnnBnnC)=5, then n(A uu B nn C)' is equal to

If n(AuuBuuC)+n(AuuBuuC)'=50,n(A)=10,n(B)=15,n(C)=20,n(AnnB)=5,n(BnnC)=6.n(CnnA)=10and n(AnnBnnC)=5,then n(AuuBuuC)' is equal to

Simplify each of the following: (a)\ (7^n-3\ xx\ 7^(n+1))/(20\ xx\ 7^n-2\ xx\ 7^n) (b)\ (5^n-6\ xx\ 5^(n+1))/(9\ xx\ 5^n-2^2xx\ 5^n) (c)\ (16\ xx\ 2^(n+1)-4\ xx\ 2^n)/(16\ xx\ 2^(n+2)-2\ xx\ 2^(n+2))

Let A = {7n:n in N} and B = {2^(3n)-1: n in N} , then

if A= {5^(n)-4n-1: n in N} and B={16(n-1):n in N} then

If the total number of relations form set A to set B is 1024, and n(A)gt1, n(B)gt1, then (A) n(A)=2, n(B)=5 (B) n(A)=5, n(B)=2 (C) n(A)=10, n(B)=1 (D) n(A)=1, n(B)=10

10^n+3(4^(n+2))+5 is divisible by (n in N)