Question

Which among the following is larger between ${99^{50}} + {100^{50}}{\text{ and 10}}{{\text{1}}^{50}}$.

Answer 1

Hint: There can be two ways to solve this problem the first one will be the direct use of calculator but however let’s come to the second method, use the binomial expansion for entities in the form of ${\left( {x + a} \right)^n}$. Convert the given number as additive or subtractive of 100. Then use binomial expansion to figure out which one is larger.

Complete step-by-step answer:
We have to find out which one is larger
$\left( 1 \right){\left( {99} \right)^{50}} + {\left( {100} \right)^{50}}$
$\left( 2 \right){\left( {101} \right)^{50}}$
Now according to Binomial Theorem the expansion of ${\left( {x + a} \right)^n}$ is given as
$\Rightarrow {\left( {x + a} \right)^n} = 1.{x^n}.{a^0} + n.{x^{n - 1}}.{a^1} + \dfrac{{n\left( {n - 1} \right)}}{{2!}}.{x^{n - 2}}.{a^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}.{x^{n - 3}}.{a^3} + .....................$
Therefore ${\left( {101} \right)^{50}} = {\left( {100 + 1} \right)^{50}}$ so expand it according to above property we have,
$\Rightarrow {\left( {100 + 1} \right)^{50}} = 1.{\left( {100} \right)^{50}}{.1^0} + 50.{\left( {100} \right)^{49}}{.1^1} + \dfrac{{50\left( {49} \right)}}{{2!}}.{\left( {100} \right)^{48}}{.1^2} + \dfrac{{50\left( {49} \right)\left( {48} \right)}}{{3!}}.{\left( {100} \right)^{47}}{.1^3} + ....{\text{ }}.........\left( 1 \right)$
Now ${\left( {99} \right)^{50}} = {\left( {100 - 1} \right)^{50}}$ so expand it according to above property we have,
$\Rightarrow {\left( {100 - 1} \right)^{50}} = 1.{\left( {100} \right)^{50}}.{\left( { - 1} \right)^0} + 50.{\left( {100} \right)^{49}}.{\left( { - 1} \right)^1} + \dfrac{{50\left( {49} \right)}}{{2!}}.{\left( {100} \right)^{48}}.{\left( { - 1} \right)^2} + \dfrac{{50\left( {49} \right)\left( {48} \right)}}{{3!}}.{\left( {100} \right)^{47}}.{\left( { - 1} \right)^3} + ........$
$\Rightarrow {\left( {100 - 1} \right)^{50}} = 1.{\left( {100} \right)^{50}}{.1^0} - 50.{\left( {100} \right)^{49}}{.1^1} + \dfrac{{50\left( {49} \right)}}{{2!}}.{\left( {100} \right)^{48}}{.1^2} - \dfrac{{50\left( {49} \right)\left( {48} \right)}}{{3!}}.{\left( {100} \right)^{47}}{.1^3} + .......{\text{ }}......\left( 2 \right)$
Now subtract equation (2) from equation (1) we have,
${\left( {100 + 1} \right)^{50}} = 1.{\left( {100} \right)^{50}}{.1^0} + 50.{\left( {100} \right)^{49}}{.1^1} + \dfrac{{50\left( {49} \right)}}{{2!}}.{\left( {100} \right)^{48}}{.1^2} + \dfrac{{50\left( {49} \right)\left( {48} \right)}}{{3!}}.{\left( {100} \right)^{47}}{.1^3} + ....$
${\left( {100 - 1} \right)^{50}} = 1.{\left( {100} \right)^{50}}{.1^0} - 50.{\left( {100} \right)^{49}}{.1^1} + \dfrac{{50\left( {49} \right)}}{{2!}}.{\left( {100} \right)^{48}}{.1^2} - \dfrac{{50\left( {49} \right)\left( {48} \right)}}{{3!}}.{\left( {100} \right)^{47}}{.1^3} + .......$
$\Rightarrow {\left( {101} \right)^{50}} - {\left( {99} \right)^{50}} = 2\left[ {\left( {50} \right){{\left( {100} \right)}^{49}} + \dfrac{{50\left( {49} \right)\left( {48} \right)}}{{3!}}{{\left( {100} \right)}^{47}} + ..........} \right]$
Multiply by 2 inside the R.H.S we have,
$\Rightarrow {\left( {101} \right)^{50}} - {\left( {99} \right)^{50}} = \left[ {\left( {100} \right){{\left( {100} \right)}^{49}} + \dfrac{{100\left( {49} \right)\left( {48} \right)}}{{3!}}{{\left( {100} \right)}^{47}} + ..........} \right]$
$\Rightarrow {\left( {101} \right)^{50}} - {\left( {99} \right)^{50}} = \left[ {{{\left( {100} \right)}^{50}} + \dfrac{{100\left( {49} \right)\left( {48} \right)}}{{3!}}{{\left( {100} \right)}^{47}} + ..........} \right]$
Now as we see that the R.H.S part of the above equation is greater than ${\left( {100} \right)^{50}}$ as the first term of the R.H.S part is ${\left( {100} \right)^{50}}$ and the remaining term is all in positive sign therefore,
$\Rightarrow {\left( {101} \right)^{50}} - {\left( {99} \right)^{50}} > {\left( {100} \right)^{50}}$
$\Rightarrow {\left( {101} \right)^{50}} > {\left( {100} \right)^{50}} + {\left( {99} \right)^{50}}$
So ${\left( {101} \right)^{50}}$ is larger.
So this is the required answer.

Note: This type of problem is one its kind, usage of binomial expansion along with the general mathematics is only helpful to solve problems of this type. It is advisable to have a good grasp over the direct formula of the binomial expansion.