Question

If $a = {99^{50}} + {100^{50}}$ and $b = {101^{50}}$ , then :
a.a < b
b.a = b
c.a > b
d.a – b = ${100^{49}}$

Answer 1

Hint: we are given that $a = {99^{50}} + {100^{50}}$ and $b = {101^{50}}$and they can be written as ${101^{50}} = {(100 + 1)^{50}}$and ${99^{50}} = {(100 - 1)^{50}}$. And expanding these using the binomial expansions ${(1 + x)^n} = 1 + {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + {}^n{C_3}{x^{n - 3}} + ......$ and${(1 - x)^n} = 1 - {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} - {}^n{C_3}{x^{n - 3}} - ......$and subtracting we get that and rearranging it gives us the required answer.

Complete step-by-step answer:
We are given that $a = {99^{50}} + {100^{50}}$ and $b = {101^{50}}$
Now we can write
$ \Rightarrow {101^{50}} = {(100 + 1)^{50}}$
Using binomial expansion ,
${(1 + x)^n} = 1 + {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + {}^n{C_3}{x^{n - 3}} + ......$
We can write the above number as
$
   \Rightarrow {(1 + 100)^{50}} = 1 + {}^{50}{C_1}{(100)^{49}} + {}^{50}{C_2}{(100)^{48}} + {}^{50}{C_3}{(100)^{47}} + ...... \\
   \Rightarrow {(1 + 100)^{50}} = 1 + 50*{(100)^{49}} + {}^{50}{C_2}{(100)^{48}} + {}^{50}{C_3}{(100)^{47}} + ...... \\
    \\
$
 Let this be equation (1)
 Now we can write
$ \Rightarrow {99^{50}} = {(100 - 1)^{50}}$
Using binomial expansion ,
${(1 - x)^n} = 1 - {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} - {}^n{C_3}{x^{n - 3}} - ......$
We can write the above number as
$
   \Rightarrow {(1 - 100)^{50}} = 1 - {}^{50}{C_1}{(100)^{49}} + {}^{50}{C_2}{(100)^{48}} - {}^{50}{C_3}{(100)^{47}} + ...... \\
   \Rightarrow {(1 - 100)^{50}} = 1 - 50*{(100)^{49}} + {}^{50}{C_2}{(100)^{48}} - {}^{50}{C_3}{(100)^{47}} + ...... \\
    \\
$
Let this be equation (2)
Let's subtract equation (2) from (1)
$
   \Rightarrow {101^{50}} - {99^{50}} = \left( {1 + 50*{{(100)}^{49}} + {}^{50}{C_2}{{(100)}^{48}} + {}^{50}{C_3}{{(100)}^{47}} + ......} \right) - \left( {1 - 50*{{(100)}^{49}} + {}^{50}{C_2}{{(100)}^{48}} - {}^{50}{C_3}{{(100)}^{47}} + ......} \right) \\
   \Rightarrow {101^{50}} - {99^{50}} = 2\left\{ {50*{{100}^{49}} + {}^{50}{C_3}{{(100)}^{47}} + ....} \right\} \\
$From this we get that
$
   \Rightarrow {101^{50}} - {99^{50}} > {100^{50}} \\
   \Rightarrow {101^{50}} > {100^{50}} + {99^{50}} \\
   \Rightarrow b > a \\
$
Therefore the correct option is a.

Note: 1.There are n+1 terms in the expansion of ${(x + y)^n}$
2.The degree of each term is n
3.The powers on x begin with n and decrease to 0
4.The powers on y begin with 0 and increase to n
5.The coefficients are symmetric