Can we compare them III?

Algebra Level 5

A = 1 + 1 4 + 1 ⋅ 4 4 ⋅ 8 + 1 ⋅ 4 ⋅ 7 4 ⋅ 8 ⋅ 12 + 1 ⋅ 4 ⋅ 7 ⋅ 10 4 ⋅ 8 ⋅ 12 ⋅ 16 + … B = 1 + 2 6 + 2 ⋅ 5 6 ⋅ 12 + 2 ⋅ 5 ⋅ 8 6 ⋅ 12 ⋅ 18 + 2 ⋅ 5 ⋅ 8 ⋅ 11 6 ⋅ 12 ⋅ 18 ⋅ 24 + … \begin{aligned} A&=&1+\dfrac{1}{4}+\dfrac{1\cdot 4}{4\cdot 8}+\dfrac{1\cdot 4 \cdot 7}{4 \cdot 8 \cdot 12}+ \dfrac{1 \cdot 4 \cdot 7 \cdot 10}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \\ B&=&1+\dfrac{2}{6}+\dfrac{2 \cdot 5}{6 \cdot 12}+\dfrac{2 \cdot 5 \cdot 8}{6 \cdot 12 \cdot 18}+\dfrac{2 \cdot 5 \cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24}+\ldots \end{aligned}

Find the relationship between A A and B B .

A > B A>B A = B A=B A < B A<B

Rohit Udaiwal by Rohit Udaiwal , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rajen Kapur
Nov 26, 2015

A = ( 1 − 3 4 ) − 1 3 = ( 1 4 ) − 1 3 = 4 1 3 A = (1 - \frac{3}{4})^\frac{-1}{3}=(\frac{1}{4})^\frac{-1}{3} = 4^\frac{1}{3} and B = ( 1 − 3 6 ) − 2 3 = ( 1 2 ) − 2 3 = 2 2 3 B =(1 - \frac{3}{6})^\frac{-2}{3}= (\frac{1}{2})^\frac{-2}{3} = 2^\frac{2}{3} . Hence A= B.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you please explain more ?

Akshat Sharda - 5 years, 6 months ago

Log in to reply

The expansion in powers of x of ( 1 + x ) n = 1 + n x + n ( n − 1 ) 1.2 x 2 + n ( n − 1 ) ( n − 2 ) 1.2.3 x 3 + … (1 + x)^n = 1 + nx + \frac{n(n-1)}{1.2}x^2 + \frac{n(n - 1)(n - 2)}{1.2.3} x^3 +\ldots is used here, called BINOMIAL THEOREM.

Rajen Kapur - 5 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...