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Chapter 4: Q. 10 (page 325)

Show that ${鈭�}_{k = 3}^{9} \frac{k}{k + 1} is equal to {鈭�}_{k = 4}^{10} \frac{k - 1}{k}$ by writing out the terms in each sum.

Short Answer

Expert verified

It is shown that ${鈭�}_{k = 3}^{9} \frac{k}{k + 1} = {鈭�}_{k = 4}^{10} \frac{k - 1}{k}$

Step by step solution

01

Step 1. Given Information

The given summations are ${鈭�}_{k = 3}^{9} \frac{k}{k + 1} a n d {鈭�}_{k = 4}^{10} \frac{k - 1}{k}$

02

Step 2. Explanation

The first sum can be written as follows,

${鈭�}_{k = 3}^{9} \frac{k}{k + 1} = \frac{3}{4} + \frac{4}{5} + \frac{5}{6} + \frac{6}{7} + \frac{7}{8} + \frac{8}{9} + \frac{9}{10}$

and the second sum can be written as,

${鈭�}_{k = 4}^{10} \frac{k - 1}{k} = \frac{3}{4} + \frac{4}{5} + \frac{5}{6} + \frac{6}{7} + \frac{7}{8} + \frac{8}{9} + \frac{9}{10}$

Hence, we can say that both the summations are equal.

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