C-13-Q118mediumsingle_mcq

The sum of the first $8$ terms of $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$ is —

a $\dfrac{255}{128}$
b $\dfrac{127}{64}$
c $\dfrac{511}{256}$
d $\dfrac{63}{32}$

ব্যাখ্যা

With $a=1$ , $r=\tfrac12$ , $n=8$ : $S_8=\dfrac{1-\left(\frac12\right)^{8}}{1-\frac12}=2\left(1-\dfrac{1}{256}\right)=2\cdot\dfrac{255}{256}=\dfrac{255}{128}$ .

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The sum of the first $8$ terms of $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$ is —

\dfrac{255}{128}

\dfrac{127}{64}

\dfrac{511}{256}

\dfrac{63}{32}

ব্যাখ্যা

With $a=1$ , $r=\tfrac12$ , $n=8$ : $S_8=\dfrac{1-\left(\frac12\right)^{8}}{1-\frac12}=2\left(1-\dfrac{1}{256}\right)=2\cdot\dfrac{255}{256}=\dfrac{255}{128}$ .

The sum of the first 888 terms of 1+12+14+18+⋯1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots1+21​+41​+81​+⋯ is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The sum of the first 888 terms of 1+12+14+18+⋯1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots1+21​+41​+81​+⋯ is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The sum of the first $8$ terms of $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$ is —

The sum of the first $8$ terms of $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$ is —