Which one is greater (A or B)?

Question

Q. If $$A=(99)^{50} + (100)^{50}$$ and $$ B=(101)^{50}$$ then-

$(a) A>B$ $(b) A<B$ $(c) A=B$

My attempt - I thought of using the binomial approximation to A and B , which gives - $$ A= (100)^{50} + (100)^{50} (\frac{50}{100}) $$ and $$ B= (100)^{50} (1+\frac{50}{100})$$ Therefore, $$ A = (100)^{50} (\frac{3}{2})$$ and $$B = (100)^{50} (\frac{3}{2})$$ Which results in equality between them , but it is not the correct answer!Though it is a simple question but Where am I doing wrong? Is it the approximation ? If it is , then why? And how to solve it ?

Any help would be appreciated

Answer 1

You need the next level in the approximation. We have $1.01^{50} \gt 1+{50 \choose 1}1^{49}\cdot 0.01+{50 \choose 2}1^{48}\cdot 0.01^2=1+0.5+25\cdot 49 \cdot 0.01^2= 1.6225$
$0.99^{50}\lt 1-{50 \choose 1}1^{49}\cdot 0.01+{50 \choose 2}1^{48}\cdot 0.01^2=1-0.5+25\cdot 49 \cdot 0.01^2= 0.6225$

Then $\frac A{100^{50}} \lt 1.6225 \lt \frac B{100^{50}}$

Answer 2

$99=100-1$

$101=100+1$

Use the Binomial Theorem:

$$\displaystyle 99^{50}=(100-1)^{50}=\sum_{n=0}^{50}(-1)^{50-n}\dbinom{50}{n}100^n $$

$$\displaystyle 101^{50}=(100+1)^{50}=\sum_{n=0}^{50}\dbinom{50}{n}100^n $$

Answer 3

Write \begin{align} 101^{50} & = (100 + 1)^{50} = 100^{50} + \binom{50}{1} \times 100^{49} + \cdots + \binom{50}{49}\times 100 + 1 \\ 99^{50} & = (100 - 1)^{50} = 100^{50} - \binom{50}{1} \times 100^{49} + \cdots - \binom{50}{49}\times 100 + 1 \\ \end{align}

Therefore, $$101^{50} - 99^{50} = 2 \times \binom{50}{1} \times 100^{49} + \cdots + 2 \times \binom{50}{49}\times 100 > 2 \times \binom{50}{1} \times 100^{49} = 100^{50}.$$

This gives you $B > A$.