Binomial Theorem 2 Question 1
1. Let
(2019 Main, 11 Jan II)
(a) 12.25
(b) 12.50
(c) 12.00
(d) 12.75
Show Answer
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
