A) \[\frac{1}{2}\left( {{3}^{50}}-1 \right)\]
B) \[\frac{1}{2}\left( {{2}^{50}}+1 \right)\]
C) \[\frac{1}{2}\left( {{3}^{50}}+1 \right)\]
D) \[\frac{1}{2}\left( {{3}^{50}} \right)\]
Correct Answer: C
Solution :
\[{{\left( 1-2\sqrt{x} \right)}^{50}}{{=}^{50}}{{C}_{0}}{{-}^{50}}{{C}_{1}}\left( 2\sqrt{x} \right){{+}^{50}}{{C}_{2}}{{\left( 2\sqrt{x} \right)}^{2}}{{-}^{50}}{{C}_{3}}{{\left( 2\sqrt{x} \right)}^{3}}+....\]\[\frac{{{\left( 1+2\sqrt{x} \right)}^{50}}{{=}^{50}}{{C}_{0}}{{+}^{50}}{{C}_{1}}\left( 2\sqrt{x} \right){{+}^{50}}{{C}_{2}}{{\left( 2\sqrt{x} \right)}^{2}}{{-}^{50}}{{C}_{3}}{{\left( 2\sqrt{x} \right)}^{3}}+....}{{{\left( 1-2\sqrt{x} \right)}^{50}}+{{\left( 1+2\sqrt{x} \right)}^{50}}=2\left[ ^{50}{{C}_{0}}{{+}^{50}}{{C}_{2}}{{2}^{2}}x{{+}^{50}}{{C}_{4}}{{2}^{3}}{{x}^{2}}+.... \right]}\]The required sum is obtained by putting x = 1 \[\frac{1+{{3}^{50}}}{2}.\]above as
You need to login to perform this action.
You will be redirected in
3 sec
