Which is larger: 99^{50} + 100^{50} or 101^{5 | vedclass

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(C) Using the Binomial Theorem,we expand $101^{50}$ and $99^{50}$ around $100$:$101^{50} = (100 + 1)^{50} = 100^{50} + 50 	imes 100^{49} + \frac{50 \

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$101^{50}$

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(C) Using the Binomial Theorem,we expand $101^{50}$ and $99^{50}$ around $100$:$101^{50} = (100 + 1)^{50} = 100^{50} + 50 	imes 100^{49} + \frac{50 	imes 49}{2} 	imes 100^{48} + \dots$ $(i)$$99^{50} = (100 - 1)^{50} = 100^{50} - 50 	imes 100^{49} + \frac{50 	imes 49}{2} 	imes 100^{48} - \dots$ $(ii)$Subtracting $(ii)$ from $(i)$:$101^{50} - 99^{50} = 2 	imes (50 	imes 100^{49} + \frac{50 	imes 49 	imes 48}{6} 	imes 100^{47} + \dots)$Since the right side is clearly greater than $100^{50}$,we have:$101^{50} - 99^{50} > 100^{50}$Therefore,$101^{50} > 100^{50} + 99^{50}$.

Which is larger: $99^{50} + 100^{50}$ or $101^{50}$?

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