Binomial Theorem 1 Question 29
31. The larger of $99^{50}+100^{50}$ and $101^{50}$ is … .
Analytical & Descriptive Questions
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Solution:
- Consider, $(101)^{50}-(99)^{50}-(100)^{50}$
$$ \begin{aligned} = & (100+1)^{50}-(100-1)^{50}-(100)^{50} \\ = & \left.{(100)^{50}(1+0.01)^{50}-(1-0.01)^{50}-1\right) } \\ = & (100)^{50}{2 \cdot\left[{ }^{50} C _1(0.01)+{ }^{50} C _3(0.01)^{3}+\ldots\right]-1 } \\ = & (100)^{50}{2\left[{ }^{50} C _3(0.01)^{3}+{ }^{50} C _5(0.01)^{5}+\ldots\right] } \\ & (101)^{50}-{(99)^{50}+(100)^{50} }>0 \\ \Rightarrow \quad & (101)^{50}>(99)^{50}+(100)^{50} \end{aligned} $$
