Can We Cancel Out The Powers?

We want to find the mathematical symbol satisfying the inequation/equation below.

$101^{50} \stackrel{?}{=} 100^{50} + 99^{50}$

Notice that $99 = 100 - 1$ and $101 = 100 + 1$ , by binomial theorem expansions , we can show that for $a>b>0$ , and $n> 0$ , all the terms of $(a+b)^n - (a-b)^n$ is strictly positive. In this case, $a= 100,b = 1$ .

$\begin{aligned} 101^{50} - 99^{50} &=& (100 + 1)^{50} - (100-1)^{50} \\ &=& \left (100^{50}+ 50\cdot 100^{49} + \dbinom{50}2 \cdot 100^{48} + \cdots + \dbinom{50}{49} \cdot100^1 + 1 \right) \\ & \phantom0 & - \left (100^{50} - 50\cdot 100^{49} + \dbinom{50}2 \cdot 100^{48} - \cdots - \dbinom{50}{49} \cdot100^1 + 1 \right) \\ &=& \left (\cancel{100^{50}}+ 50\cdot 100^{49} + \cancel{\dbinom{50}2 \cdot 100^{48}} + \cdots + \cancel{\dbinom{50}{49} \cdot100^1} + 1 \right) \\ & \phantom0 & - \left (\cancel{100^{50}} - 50\cdot 100^{49} + \cancel{\dbinom{50}2 \cdot 100^{48}} - \cdots - \cancel{\dbinom{50}{49} \cdot100^1} + 1 \right) \\ &=& 2\left [50\cdot 100^{49} + \cdots + 1 \right ] \\ &=& 2\left [50\cdot 100^{49} + (\text{sum of positive numbers}) \right ] \\ &> & 2\left [50\cdot 100^{49} \right ] = 100 \cdot 100^{49} = 100^{49 + 1} = 100^{50} \\ \end{aligned}$

Thus, we have $101^{50} - 99^{50} > 100^{50}$ , or equivalently, $101^{50} > 100^{50} + 99^{50}$ , this inequality tells us that $101^{50}$ is larger than $100^{50} + 99^{50}$ .

Claim: $1 < 1.01 ^ {50} - 0.99 ^ { 50}$ .

We use the binomial theorem: $(1+x)^{50} = 1 + { 50 \choose 1 } x + { 50 \choose 2 } x^2 + { 50 \choose 3 } x^3 + \ldots$

With $x = 0.01$ , we see that $1.01^{50} = 1 + {50\choose 1} \times 0.01 +{50\choose 2} \times 0.01^2 + {50\choose 3} \times 0.01^3 + \ldots$ .
With $x = -0.01$ , we see that $0.99 ^ {50} = 1 - {50\choose 1} \times 0.01 + {50\choose 2} \times 0.01^2 - {50\choose 3} \times 0.01^3 + \ldots$ .

Subtracting these 2 equations, we get that

$1.01^{50} - 0.99 ^ { 50} = 2 \times {50\choose 1} \times 0.01 + 2 \times {50 \choose 3} \times 0.01^3 + 2 \times { 50 \choose 5 } \times 0.01^5 \ldots > 1 + \ldots > 1$ .

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Multiplying throughout by $100^{50}$ , we see that $99 ^{50} + 100 ^ {50} < 101 ^ {50}$ .

We can determine which number is the larger of the two by dividing one by the other. Recall that $\frac {a}{b}<1$ if $a<b$ .

Using the same principle:

$\frac {100^{50}+99^{50}}{101^{50}}=\big(\frac {100}{101}\big)^{50}+\big(\frac {99}{101}\big)^{50}<1$

Since $100<101$ and $99<101$ , we know these two fractions are less than 1, respectively. Furthermore, we can be certain the summation of the two fractions is also less than 1 because $\lim_{x\to\infty} a^x\rightarrow0$ if $a<1$ (and 50 is a relatively large power). Therefore, our denominator must be larger than our numerator: $101^{50}>100^{50}+99^{50}$ .

101^50=(100+1)^50 =100^50+100×100^49+....+1 =100^50+100^50+...+1 So

101^50 > 100^50+99^50

We suppose 101^50>100^50+99^50. Dividing for 99^50 we obtain (101/99)^50-(100/99)^50>1 that is true so 101^50>100^50+99^50.

I used the Log and Antilog table to find the approx value of both the terms and then compared them.

First of all, we will find the approx value of $101^{50}$ ,

So,

$\log 101^{50}=\log x \\ 50 \log 101=\log x \\ \Rightarrow \log 101=2.0043 \\ 50 \times 2.0043=\log x \\ 100.215=\log x$

Now we will find the Antilog of this,

$x =$ antilog $0.215$

(with $100$ as the characteristic)

We get value as $1641$ and after putting the characteristic we get the approx value as,

$1641000 \ldots 000$

(with total $101$ digits)

Now we will find the approx value of $99^{50}$ with the same procedure which comes out to be

$6026000 \ldots 000$

(with total $100$ digits)

And $100^{50} = 100000 \ldots 0000$

(with $101$ digits )

So the approx value of $2^{nd}$ expression is

$16026000\ldots 000$

(with $101$ digits)

So by comparing the approx value of both,

$\boxed{101^{50} > 100^{50} + 99^{50}}$

100^(50) + 99^(50) < 100^(50) + 100^(50) = 2x(100^(50)) < 100x(100^(50)) =100^(51) More precisely 100^(50) + 99^(50) is much less then 2% of 100^(51); so way smaller.

We can rule out the equality by Fermat's Last Theorem.