A) \[{{2}^{100}}\]
B) \[{{2}^{99}}\]
C) 202
D) 200
Correct Answer: A
Solution :
Here,\[^{101}{{C}_{1}}{{+}^{101}}{{C}_{2}}{{S}_{1}}+...{{+}^{101}}{{C}_{101}}{{S}_{100}}=\alpha {{T}_{100}}\] \[\Rightarrow \]\[^{101}{{C}_{1}}{{+}^{101}}{{C}_{2}}(1+q){{+}^{101}}{{C}_{3}}(1+q+{{q}^{2}})+\]\[...{{+}^{101}}{{C}_{101}}(1+q+...+{{q}^{100}})\] \[=2\alpha \frac{\left( 1-{{\left( \frac{1+q}{2} \right)}^{101}} \right)}{(1-q)}\] \[\Rightarrow \]\[^{101}{{C}_{1}}(1-q){{+}^{101}}{{C}_{2}}(1-{{q}^{2}})+...\]\[{{+}^{101}}{{C}_{101}}(1-{{q}^{101}})\] \[=2\alpha \left( 1-{{\left( \frac{1+q}{2} \right)}^{101}} \right)\] \[\Rightarrow \]\[({{2}^{101}}-1)-[{{(1+q)}^{101}}-1]\] \[=2\alpha \left( 1-{{\left( \frac{1+q}{2} \right)}^{101}} \right)\] \[\Rightarrow \]\[={{2}^{101}}\left( 1-{{\left( \frac{1+q}{2} \right)}^{101}} \right)=2\alpha \left( 1-{{\left( \frac{1+q}{2} \right)}^{101}} \right)\] \[\Rightarrow \]\[\alpha ={{2}^{100}}\]
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