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Geometric Progressions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSGeometric Progressions3 Marks
Question
Find the sum: $\sum\limits_{\text{n}=1}^{10}\bigg\{\Big(\frac12\Big)^{\text{n}-1}+\Big(\frac15\Big)^{\text{n}+1}\bigg\}.$
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Answer

$\sum\limits_{\text{n}=1}^{10}\bigg\{\Big(\frac12\Big)^{\text{n}-1}+\Big(\frac15\Big)^{\text{n}+1}\bigg\}$
$=\sum\limits_{\text{n}=1}^{10}\Big(\frac12\Big)^{\text{n}-1}+\sum\limits_{\text{n}=1}^{10}\Big(\frac15\Big)^{\text{n}+1}$
$=1+\frac12+\frac{1}{2^2}+\ \cdots\ +\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\ \cdots$
$=\frac{\Big(1-\frac{1}{2^{10}}\Big)}{1-\frac12}+\frac{\frac15\Big(1-\frac{1}{5^{10}}\Big)}{1-\frac15}$
$=\frac{2^{10}-1}{2^9}+\frac{5^{10}-1}{5^{11}}$

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