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) $12.75$
Show Answer
Answer:
Correct Answer: 1. (a)
Solution:
- We have,
$ (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[{ }^{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] $
$= 2\left[{ }^{50} C_{0} x^{50}+{ }^{50} C_{2} x^{48} \cdot 10^{2}+{ }^{50} C_{4} x^{46} \cdot 10^{4}\right. $ $ +\ldots+{ }^{50} C_{50} \cdot 10^{50}]$
By comparing coefficients, we get
$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}} $
$[\therefore { }^{50} C_{48} = { }^{50} C_{2}]$
$\quad=\frac{50 \times 49}{2 \cdot(10 \times 10)}=\frac{5 \times 49}{20}=\frac{245}{20}=12.25$
