A) \[\frac{{{(1+\alpha )}^{50}}(50-\alpha )+{{\alpha }^{51}}}{{{\alpha }^{49}}}\]
B) \[\frac{{{(1+\alpha )}^{51}}(50-\alpha )+{{\alpha }^{2}}}{{{\alpha }^{50}}}\]
C) \[{{\left( \frac{1+\alpha }{\alpha } \right)}^{50}}+{{\alpha }^{2}}\]
D) \[\left( \frac{50-\alpha }{\alpha } \right)+{{\alpha }^{2}}\]
Correct Answer: A
Solution :
Let \[S=1+2\,\left( 1+\frac{1}{\alpha } \right)+3\,{{\left( 1+\frac{1}{\alpha } \right)}^{2}}\] \[...+50\,{{\left( 1+\frac{1}{\alpha } \right)}^{49}}\] \[S\left( 1+\frac{1}{\alpha } \right)=1\left( 1+\frac{1}{\alpha } \right)+2\,{{\left( 1+\frac{1}{\alpha } \right)}^{2}}\] \[...+50\,{{\left( 1+\frac{1}{\alpha } \right)}^{50}}\] ?(ii) On subtracting Eq. (ii) from Eq. (i), we get \[-\frac{S}{\alpha }=1+\left( 1+\frac{1}{\alpha } \right)+{{\left( 1+\frac{1}{\alpha } \right)}^{2}}\] \[+...+\,\left( 1+\frac{1}{\alpha } \right)+{{\left( 1+\frac{1}{\alpha } \right)}^{2}}\] \[\Rightarrow \] \[50{{\left( 1+\frac{1}{\alpha } \right)}^{50}}-\frac{S}{\alpha }=\frac{{{\left( 1+\frac{1}{\alpha } \right)}^{50}}-1}{1+\frac{1}{\alpha }-1}\] \[\Rightarrow \] \[50\,\frac{{{(\alpha +1)}^{50}}}{{{\alpha }^{50}}}-\frac{S}{\alpha }=\frac{{{(1+\alpha )}^{50}}-{{\alpha }^{50}}}{{{\alpha }^{49}}}\] \[\Rightarrow \] \[50\,{{(\alpha +1)}^{50}}-S{{\alpha }^{49}}=\alpha {{(1+\alpha )}^{50}}-{{\alpha }^{51}}\] \[\Rightarrow \] \[-S=\frac{{{(1+\alpha )}^{50}}-(50+\alpha )-{{\alpha }^{51}}}{{{\alpha }^{49}}}\] \[\Rightarrow \] \[S=\frac{{{(1+\alpha )}^{50}}\,(50-\alpha )+{{\alpha }^{51}}}{{{\alpha }^{49}}}\]
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