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: According to given in the question we have to determine the smaller of ${99^{100}} + {100^{100}}$and${101^{100}}$. So, first of all we have to rearrange the term ${101^{100}}$in the binomial form so, that we can apply the binomial theorem.
Now, we have to use the binomial theorem formula to expand the term and the formula for the binomial theorem is as mentioned below:

Formula used:
$ \Rightarrow {(a + b)^n} = {a^n} + (c_1^n){a^{n - 1}}b + (c_2^n){a^{n - 2}}{b^2} + ................. + (c_{n - 1}^n)a{b^{n - 1}} + b...............(A)$
Hence, with the help of the formula (A) above, we can expand the term in the form of binomial expression.
Now, we have to solve the expression obtained and we have to let any integer for the expansion.
Now, we have to check for equality after solving the expansion and then check for the correct option.

Complete step-by-step answer:
Step 1: First of all we have to rearrange the term ${101^{100}}$in the binomial form so, that we can apply the binomial theorem as mentioned in the solution hint. Hence,
$ \Rightarrow {(101)^{100}} = {(100 + 1)^{100}}..............(1)$
Step 2: Now, we have to apply the binomial theorem formula (A) which is as mentioned in the solution hint in the expression (1) which is as obtained in the solution step 1. Hence,
$ \Rightarrow {(100 + 1)^{100}} = \left( {c_0^{100}} \right){100^{100}} + \left( {c_1^{100}} \right){100^{99}} + \left( {c_2^{100}} \right){100^{98}} + ......... + \left( {c_{100}^{100}} \right){100^0}$…………(2)
Step 3: Now, we have to solve the expression as obtained on the solution step (3) hence,
$ \Rightarrow {100^{100}} + 100 \times {100^{99}} + .....(....)$
Now, we have to let the remaining terms of the expression above as k. Hence,
$ \Rightarrow {100^{100}} + {100^{100}} + k,$
Where, $k > 0$
Step 4: Now, from the expression as obtained in the solution step 3 we can say that,
$
    \Rightarrow {100^{100}} + {100^{100}} + k > {100^{100}} + {99^{100}} \\
    \Rightarrow {101^{100}} > {100^{100}} + {99^{100}} \\
 $

Final solution: Hence, with the help of the formula (A) of binomial theorem we have determined that ${101^{100}} > {100^{100}} + {99^{100}}$. Therefore option (C) is correct.

Note:
To expand the term ${101^{100}}$in the form of binomial expansion it is necessary that first of all we have to convert the term ${101^{100}}$ in the form of ${(100 + 1)^{100}}$ so that we can apply the formula (A) which is mentioned in the solution hint.
For the remaining terms of the expansion it is necessary that we have to let the remaining expression as any constant term as we let as k.