Question

Which of the following is larger? 99^50 + 100^50 or 101^50

Answer 1

Answer:

101^50 is greater than 99^50 + 100^50

Step-by-step explanation:

Showing that 101^50 > 99^50 + 100^50 is equivalent to showing that

101^50 - 99^50 > 100^50.

Since 101 and 99 differ from 100 by 1, the binomial theorem will expand nicely around 100 for these values.

$101^{50} = (100+1)^{50} = \sum_{k=0}^{50}\left(\begin{array}{c}50\\ k\end{array}\right)100^k$

and

$99^{50} = (100-1)^{50} = \sum_{k=0}^{50}(-1)^k\left(\begin{array}{c}50\\ k\end{array}\right)100^k$

Taking the difference, we have

$101^{50} - 99^{50}= 2\left(\begin{array}{c}50\\ 1\end{array}\right)100 + 2\left(\begin{array}{c}50\\ 3\end{array}\right)100^3 +\dots+2\left(\begin{array}{c}50\\ 49\end{array}\right)100^{49}$

Just taking the last term of this expansion is enough to give the result, for then

$101^{50} - 99^{50} > 2\left(\begin{array}{c}50\\ 49\end{array}\right)100^{49} = 2\times 50\times 100^{49} = 100^{50}$

Answer 2

101^50 - 99^50 > 100^50

Step-by-step explanation:

101^50

= (100 + 1)^50

= 100^50 + (50C1) * 100^49 + (50C2) * 100^48 + (50C3) * 100^47 + ... + (50C49) * 100^1 + 1

99^50

= (100 - 1)^50

= 100^50 - (50C1) * 100^49 + (50C2) * 100^48 - (50C3) * 100^47 + ... - (50C49) * 100^1 + 1

101^50 - 99^50

= 2 [ (50C1) * 100^49 + (50C3) * 100^47 + (50C5) * 100^45 + ... + (50C49) * 100^1 ]

= 2 [ (50 * 100^49) + (17150 * 100^47) + .... ], Computing 50C1 and 50C3, ignoring the remaining positive terms

= 2 [ (50 * 100^49) + (1.7150 * 100^49) + .... ]

= 2 [ (50 + 1.7150) * 100^49 + ... ]

= 2 [ 51.7150 * 100^49 + ... ]

= 103.43 * 100^49 + ...

= 1.0343 * 100^50 + ...

> 100^50

Thus

101^50 - 99^50 > 100^50

or

101^50 > 99^50 + 100^50

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