Sum of the first n terms of the series n (1 / 2) + (3 / 4) + (7 / 8) + (15 / 16) + ... ... .n is equal to

n (c) : Let S_n=(1 / 2)+(3 / 4)+(7 / 8)+(15 / 16)+ * s . upto n terms n =(1-(1 / 2))+(1-(1 / 2^2))+(1-(1 / 2^3))+ * s * +(1-(1 / 2^n)) n =n-((1 / 2)+(1 / 2^2)+ * s * +(1 / 2^n))=n-frac(1 / 2)(1-(1 / 2^n))1-(1 / 2)=n+2^-n-1 .n\

Sum of the first n terms of the series \frac{1}{2} + ... | Sequence & Series (Mathematics)

\begin{aligned}\n& { (c) : Let } S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots \text {. upto } n \text { terms } \\\n& =\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+\cdots \cdot+\left(1-\frac{1}{2^n}\right) \\\n& =n-\left(\frac{1}{2}+\frac{1}{2^2}+\cdots \cdot+\frac{1}{2^n}\right)=n-\frac{\frac{1}{2}\left(1-\frac{1}{2^n}\right)}{1-\frac{1}{2}}=n+2^{-n}-1 .\n\end{aligned}

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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