Advertisements
Advertisements
प्रश्न
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
उत्तर
We have to prove that `(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Let x = `(2^n+2^(n-1))/(2^(n+1)-2^n)`
`=(2^n(1+1xx2^-1))/(2^n(2^1-1))`
`=(1+1/2)/(2-1)`
`rArrx=3/2`
Hence, `(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
APPEARS IN
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
Simplify:
`(16^(-1/5))^(5/2)`
Simplify:
`root5((32)^-3)`
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
If ax = by = cz and b2 = ac, show that `y=(2zx)/(z+x)`
Write \[\left( 625 \right)^{- 1/4}\] in decimal form.
Simplify \[\left[ \left\{ \left( 625 \right)^{- 1/2} \right\}^{- 1/4} \right]^2\]
For any positive real number x, write the value of \[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\]
The simplest rationalising factor of \[\sqrt[3]{500}\] is
If x = \[\sqrt[3]{2 + \sqrt{3}}\] , then \[x^3 + \frac{1}{x^3} =\]