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  • The sum of the series <img alt="\sum_{r=0}^n(-1)^r{ }^n C_r\left(\frac{1}{2^r}+\frac{3^r}{2^{2 r}}+\frac{7^r}{2^{3 r}}+\frac{15^r}{2^{4 r}}+\ldots . . m \text { terms }\right)" src="https://en

The sum of the series \sum_{r=0}^n(-1)^r{ }^n C_r\left(\frac{1}{2^r}+\frac{3^r}{2^{2 r}}+\frac{7^r}{2^{3 r}}+\frac{15^r}{2^{4 r}}+\ldots . . m \text { terms }\right) is

Option: 1

\frac{2^{m n}-1}{2^{m n}\left(2^n-1\right)}


Option: 2

\frac{2^{m n}-1}{\left(2^n-1\right)}


Option: 3

\frac{2^{m n}+1}{\left(2^n+1\right)}


Option: 4

None of these


Answers (1)

best_answer

\\\sum_{r=0}^n(-1)^r{ }^n C_r\left(\frac{1}{2^r}+\frac{3^r}{2^{2 r}}+\frac{7^r}{2^{3 r}}+\ldots . \text { uptom terms }\right) \\ =\sum_{r=0}^n(-1)^r{ }^n C_r \cdot \frac{1}{2^r}+\sum_{r=0}^n(-1)^r \cdot{ }^n C_r \frac{3^r}{2^{2 r}} +\sum_{r=0}^n(-1)^r{ }^n C_r \frac{7^r}{2^{3 r}}+\ldots . .

=\left(1-\frac{1}{2}\right)^n+\left(1-\frac{3}{4}\right)^n+\left(1-\frac{7}{8}\right)^n+\ldots    upto m terms

=\frac{1}{2^n}+\frac{1}{4^n}+\frac{1}{8^n}+\ldots        upto m terms

=\frac{\frac{1}{2^n}\left(1-\frac{1}{2^{n m}}\right)}{\left(1-\frac{1}{2^n}\right)}=\frac{2^{m n}-1}{2^{m n}\left(2^n-1\right)}

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shivangi.shekhar

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