Show that `(101)^(50) gt (100)^(50) + (99)^(50)` .

Using binomial theorem, prove that (101)^(50)> 100^(50)+99^(50)dot

If x = (99)^(50) + (100)^(50) and y = (101)^(50) then

Let A=[(1,0,0),(1,0,1),(0,1,0)] satisfies A^(n)=A^(n-2)+A^(2)-I for n ge 3 . And trace of a square matrix X is equal to the sum of elements in its proncipal diagonal. Further consider a matrix underset(3xx3)(uu) with its column as uu_(1), uu_(2), uu_(3) such that A^(50) uu_(1)=[(1),(25),(25)], A^(50) uu_(2)=[(0),(1),(0)], A^(50) uu_(3)=[(0),(0),(1)] Then answer the following question : Trace of A^(50) equals

The value of 99^(50) - 99.98^(50) + (99*98)/(1*2) (97)^(50) -…+ 99 is

The value of ((50),(0)) ((50),(1)) + ((50), (1)) ((50),(2)) +............+ ((50),(49))((50),(50)) is

Solve cos^(50) x- sin^(50)x=1

Show that the points A(1,-2,-8), B(5,0,-2) and C(11,3,7) are collinear.

Simplify ( 41 ^(2) - 9 ^(2))/( 50)

Prove that: sum_(k=1)^(100)sin(k x)cos(101-k)x=50"sin"(101 x)

The coefficient of x^(50) in the series sum_(r=1)^(101)rx^(r-1)(1+x)^(101-r) is