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Chapter 7: Q. 42 (page 615)

Evaluate the finite sums.
${鈭�}_{j = 11}^{200} {(\frac{3}{2})}^{j}$

Short Answer

Expert verified

On evaluating, we get,

$2 ({(\frac{3}{2})}^{201} - {(\frac{3}{2})}^{11})$

Step by step solution

01

Step 1. Given information.

Consider the given question,

${鈭�}_{j = 11}^{200} {(\frac{3}{2})}^{j}$

02

Step 2. Use the sum of n terms of geometric series.

The geometric series ${鈭�}_{j = 11}^{200} {(\frac{3}{2})}^{j}$ in the expanded form is ${鈭�}_{j = 11}^{200} {(\frac{3}{2})}^{j} = {(\frac{3}{2})}^{11} + {(\frac{3}{2})}^{12} + . . . + {(\frac{3}{2})}^{200}$ .

The series has the first term $a = {(\frac{3}{2})}^{11}$ .

The common ratio $r = \frac{3}{2}$ with no. of terms,

role="math" localid="1649269759876" $n = 190$

The finite sum of the geometric series with ratio greater than $1$ , $S_{n} = \frac{a (r^{n} - 1)}{r - 1}$ .

$S_{190} = \frac{{(\frac{3}{2})}^{11} ({(\frac{3}{2})}^{190} - 1)}{(\frac{2}{3}) - 1} = \frac{2 {(\frac{3}{2})}^{11} ({(\frac{3}{2})}^{190} - 1)}{3 - 2} = 2 ({(\frac{3}{2})}^{201} - {(\frac{3}{2})}^{11})$

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Most popular questions from this chapter

Find the values of x for which the series ${鈭�}_{k = 0}^{鈭�} x^{k}$ converges.

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