Binomial Theorem 1 Question 29
31.
The larger of $99^{50}+100^{50}$ and $101^{50}$ is … .
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Answer:
Correct Answer: 31. $(101)^{50}$
Solution:
- Consider, $(101)^{50}-(99)^{50}-(100)^{50}$
$= (100+1)^{50}-(100-1)^{50}-(100)^{50} $
$= (100)^{50}(1+0.01)^{50}-(1-0.01)^{50}-1 $
$= (100)^{50}{2 \cdot\left[{ }^{50} C_{1}(0.01)+{ }^{50} C_{3}(0.01)^{3}+\ldots\right]-1} $
$= (100)^{50}{2\left[{ }^{50} C_{3}(0.01)^{3}+{ }^{50} C_{5}(0.01)^{5}+\ldots\right]} $
$\Rightarrow (101)^{50}-{(99)^{50}+(100)^{50}}>0 $
$\Rightarrow (101)^{50}>(99)^{50}+(100)^{50}$
