Question

The smaller of \[{99^{100}} + {100^{100}}\]and \[{101^{100}}\], is​
A) \[{99^{100}} + {100^{100}}\]
B) Both are equal
C) \[{101^{100}}\]
D) None of these

Answer 1

Hint: Here we will first rewrite 101 as 1+100 and then use binomial expansion to expand \[{101^{100}}\] and then finally compare the two given values to find which smaller.
In order to decide which value is smaller, we need to compare the values of the expressions. If the value of the expression turns out to be less than that of other expressions then it is smaller.

Complete step by step answer:
First we will consider the expression \[{101^{100}}\]
Now since 101 can be written as 1+100
Therefore,
\[\Rightarrow {101^{100}} = {\left( {1 + 100} \right)^{100}}\]
The general binomial expansion of \[{\left( {a + b} \right)^n}\] is given by:
\[\Rightarrow {\left( {a + b} \right)^n} = {}^n{C_0}{\left( a \right)^0}{\left( b \right)^n} + {}^n{C_1}{\left( a \right)^1}{\left( b \right)^{n - 1}} + {}^n{C_2}{\left( a \right)^2}{\left( b \right)^{n - 2}} + ..................... + {}^n{C_n}{\left( a \right)^n}{\left( b \right)^0} \\
    \\
 \]
Now using binomial expansion to expand the above expression we get:-
\[\Rightarrow {101^{100}} = {}^{100}{C_0}{\left( 1 \right)^0}{\left( {100} \right)^{100}} + {}^{100}{C_1}{\left( 1 \right)^1}{\left( {100} \right)^{99}} + {}^{100}{C_2}{\left( 1 \right)^2}{\left( {100} \right)^{98}} + ..................... + {}^{100}{C_{100}}{\left( 1 \right)^{100}}{\left( {100} \right)^0}\]
On simplifying it we get:
\[\Rightarrow {101^{100}} = {}^{100}{C_0}{\left( {100} \right)^{100}} + {}^{100}{C_1}{\left( {100} \right)^{99}} + {}^{100}{C_2}{\left( {100} \right)^{98}} + ..................... + {}^{100}{C_{100{\text{ }}}}{\text{ }}........................\left( 1 \right)\]
Now since we know that
\[\Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Therefore, applying for \[{}^{100}C0\] we get:-
\[\Rightarrow {}^{100}{C_0} = \dfrac{{100!}}{{0!\left( {100 - 0} \right)!}}\]
Simplifying it further we get:-
\[
\Rightarrow {}^{100}{C_0} = \dfrac{{100!}}{{1\left( {100!} \right)}} \\
\Rightarrow {}^{100}{C_0} = 1 \\
 \]
Similarly applying for \[{}^{100}{C_1}\] we get:-
\[\Rightarrow {}^{100}{C_1} = \dfrac{{100!}}{{1!\left( {100 - 1} \right)!}}\]
Simplifying it further we get:-
\[\Rightarrow {}^{100}{C_1} = \dfrac{{100 \times 99!}}{{1\left( {99!} \right)}} \\
\Rightarrow {}^{100}{C_1} = 100 \\
 \]
Now applying for \[{}^{100}{C_2}\] we get:-
\[{}^{100}{C_2} = \dfrac{{100!}}{{1!\left( {100 - 2} \right)!}}\]
Simplifying it further we get:-
\[\Rightarrow {}^{100}{C_2} = \dfrac{{100 \times 99 \times 98!}}{{2 \times 1 \times \left( {98!} \right)}} \\
\Rightarrow {}^{100}{C_2} = 50 \times 99 \\
\Rightarrow {}^{100}C2 = 4950 \\
 \]
Also applying for \[{}^{100}{C_{100}}\] we get:-
\[\Rightarrow {}^{100}{C_{100}} = \dfrac{{100!}}{{100!\left( {100 - 100} \right)!}}\]
Simplifying it further we get:-
\[
 \Rightarrow {}^{100}{C_{100}} = \dfrac{{100!}}{{\left( {0!} \right)\left( {100!} \right)}} \\
\Rightarrow {}^{100}{C_{100}} = 1 \\
 \]
Hence now putting in the values in equation 1 we get:-
\[\Rightarrow {101^{100}} = \left( 1 \right){\left( {100} \right)^{100}} + 100{\left( {100} \right)^{99}} + \left( {4950} \right){\left( {100} \right)^{98}} + ..................... + 1 \\
\Rightarrow {101^{100}} = {\left( {100} \right)^{100}} + 100{\left( {100} \right)^{99}} + \left( {4950} \right){\left( {100} \right)^{98}} + ..................... + 1 \\
 \]
Now it can be clearly seen that
\[\Rightarrow {101^{100}} = {\left( {100} \right)^{100}} + 100{\left( {100} \right)^{99}} + \left( {4950} \right){\left( {100} \right)^{98}} + ..................... + 1 > {99^{100}} + {100^{100}}\]
Therefore,
\[\Rightarrow {101^{100}} > {99^{100}} + {100^{100}}\]
Therefore,
\[\Rightarrow {99^{100}} + {100^{100}}\] is smaller than \[{101^{100}}\]

Hence option A is the correct option.

Note:
The binomial expression describes the algebraic expansion of powers of a binomial.
The general binomial expansion of \[{\left( {a + b} \right)^n}\] is given by:
\[{\left( {a + b} \right)^n} = {}^n{C_0}{\left( a \right)^0}{\left( b \right)^n} + {}^n{C_1}{\left( a \right)^1}{\left( b \right)^{n - 1}} + {}^n{C_2}{\left( a \right)^2}{\left( b \right)^{n - 2}} + ..................... + {}^n{C_n}{\left( a \right)^n}{\left( b \right)^0} \\
 \]
The student should not expand the whole binomial term as it can be directly which number is smaller.
If there are two numbers and then if \[n_1 > n_2\] this implies that \[n_1\]has greater value in comparison to \[n_2\] and hence \[n_1\]is greater than \[n_2\] or \[n_2\]is smaller than \[n_1\]
Similarly if \[n_2 > n_1\] this implies that \[n_2\]has greater value in comparison to \[n_1\] and hence \[n_2\]is greater than \[n_1\] or \[n_1\]is smaller than \[n_2\]