Binomial Theorem And Its Simple Applications Question 186
Question: The value of $ ( \begin{matrix} 30 \\ 0 \\ \end{matrix} )( \begin{matrix} 30 \\ 10 \\ \end{matrix} )-( \begin{matrix} 30 \\ 1 \\ \end{matrix} )( \begin{matrix} 30 \\ 11 \\ \end{matrix} )+( \begin{matrix} 30 \\ 2 \\ \end{matrix} )( \begin{matrix} 30 \\ 12 \\ \end{matrix} )… $ $ +( \begin{matrix} 30 \\ 20 \\ \end{matrix} )( \begin{matrix} 30 \\ 30 \\ \end{matrix} ) $ is where $ ( \begin{matrix} n \\ r \\ \end{matrix} )={{,}^{n}}C_{r} $
Options:
A) $ ( \begin{matrix} 30 \\ 10 \\ \end{matrix} ) $
B) $ ( \begin{matrix} 30 \\ 15 \\ \end{matrix} ) $
C) $ ( \begin{matrix} 60 \\ 30 \\ \end{matrix} ) $
D) $ ( \begin{matrix} 31 \\ 10 \\ \end{matrix} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] To find $ ^{30}C_0{{,}^{30}}C_{10}{{-}^{30}}C_1{{,}^{30}}C_{11}+{{,}^{30}}C_2{{,}^{30}}C_{12}-… $
$ +{{,}^{30}}C_{20}{{,}^{30}}C_{30} $
We know that $ {{(1+x)}^{30}}={{,}^{30}}C_0+{{,}^{30}}C_1x+{{,}^{30}}C_2x^{2} $
$ +…..+{{,}^{30}}C_{20}x^{20}+…..{{,}^{30}}C_{30}x^{30}…(1) $
$ {{(x-1)}^{30}}={{,}^{30}}C_0x^{30}-{{,}^{30}}C_1x^{29}+….+{{,}^{30}}C_{10}x^{20} $
$ {{-}^{30}}C_{11}x^{19}+{{,}^{30}}C_{12}x^{18}+…{{,}^{30}}C_{30}x^{0}…(2) $
Multiplying $ eq^{n} $ (1) and (2), we get
$ {{(x^{2}-1)}^{30}}=(,1,)\times (,2,) $
Equating the coefficients of $ x^{20} $ on both sides,
we get
$ ^{30}C_{10}={{,}^{30}}C_0{{,}^{30}}C_{10}-{{,}^{30}}C_1{{,}^{30}}C_{11}+ $
$ ^{30}C_2{{,}^{30}}C_{12}-……+{{,}^{30}}C_{20}{{,}^{30}}C_{30} $
$ \therefore $ Req. value is $ ^{30}C_{10} $
BETA
