A) \[{{K}_{sp}}\]
B) \[AgI{{O}_{3}}\]
C) 0
D) \[1.0\times {{10}^{-8}},\]
Correct Answer: B
Solution :
Use binomial theorem of expansion for \[{{(1+x)}^{n}}\]and solve it for\[x=-1.\] We know that, \[{{(1+x)}^{20}}{{=}^{20}}{{C}_{0}}{{+}^{20}}{{C}_{1}}x+....{{+}^{20}}{{C}_{10}}{{x}^{10}}+\] \[....{{+}^{20}}{{C}_{20}}{{x}^{20}}\] On putting\[x=-1\]in above expansion, we get \[0{{=}^{20}}{{C}_{0}}{{-}^{20}}{{C}_{1}}+...{{-}^{20}}{{C}_{9}}{{+}^{20}}{{C}_{10}}{{-}^{20}}{{C}_{11}}+\] \[.....{{+}^{20}}{{C}_{20}}\] \[\Rightarrow \] \[0{{=}^{20}}{{C}_{0}}{{-}^{20}}{{C}_{1}}+...{{-}^{20}}{{C}_{9}}{{+}^{20}}{{C}_{10}}\] \[{{-}^{20}}{{C}_{9}}+.....{{+}^{20}}{{C}_{0}}\] \[\Rightarrow \]\[0=2{{(}^{20}}{{C}_{0}}{{-}^{20}}{{C}_{1}}+...{{-}^{20}}{{C}_{9}}){{+}^{20}}{{C}_{10}}\] \[\Rightarrow \] \[^{20}{{C}_{10}}=2{{(}^{20}}{{C}_{0}}{{-}^{20}}{{C}_{1}}+{{...}^{20}}{{C}_{10}})\] \[\Rightarrow \]\[^{20}{{C}_{0}}{{-}^{20}}{{C}_{1}}+....{{+}^{20}}{{C}_{10}}=\frac{1}{2}{{\,}^{20}}{{C}_{10}}\]
You need to login to perform this action.
You will be redirected in
3 sec
