Binomial Theorem 2 Question 1
1. Let $(x+10)^{50}+(x-10)^{50}=a _0+a _1 x+a _2 x^{2}+\ldots+a _{50} x^{50}$, for all $x \in R$; then $\frac{a _2}{a _0}$ is equal to
(2019 Main, 11 Jan II)
(a) 12.25
(b) 12.50
(c) 12.00
(d) 12.75
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Answer:
Correct Answer: 1. (a)
Solution:
- We have,
$$ \begin{aligned} & (x+10)^{50}+(x-10)^{50}=a _0+a _1 x+a _2 x^{2}+\ldots+a _{50} x^{50} \\ \therefore & a _0+a _1 x+a _2 x^{2}+\ldots+a _{50} x^{50} \\ = & {\left[{ }^{50} C _0 x^{50}+{ }^{50} C _1 x^{49} 10+{ }^{50} C _2 x^{48} \cdot 10^{2}+\ldots+{ }^{50} C _{50} 10^{50}\right) } \\
- & {\left.\left[{ }^{50} C _0 x^{50}-{ }^{50} C _1 x^{49} 10+{ }^{50} C _2 x^{48} 10^{2}-\ldots+{ }^{50} C _{50} 10^{50}\right)\right] } \\ = & 2\left[{ }^{50} C _0 x^{50}+{ }^{50} C _2 x^{48} \cdot 10^{2}+{ }^{50} C _4 x^{46} \cdot 10^{4}\right. \\ & \left.+\ldots+{ }^{50} C _{50} \cdot 10^{50}\right] \end{aligned} $$
By comparing coefficients, we get
$$ \begin{aligned} & a _2=2{ }^{50} C _{48}(10)^{48} ; a _0=2{ }^{50} C _{50}(10)^{50}=2(10)^{50} \\ & \therefore \frac{a _2}{a _0}=\frac{2\left({ }^{50} C _2\right)(10)^{48}}{2(10)^{50}}=2 \frac{50 \cdot 49}{1 \cdot 2} \frac{(10)^{48}}{2 \cdot(10)^{50}} \\ & \quad=\frac{50 \times 49}{2 \cdot(10 \times 10)}=\frac{5 \times 49}{20}=\frac{245}{20}=12.25 \end{aligned} $$
