How do you evaluate the sum represented by $5 \sum n = 1 \frac{n}{2 n + 1}$ $sum_(n=1)^5n/(2n+1)$ ?

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How do you evaluate the sum represented by $5 \sum n = 1 \frac{n}{2 n + 1}$ $sum_(n=1)^5n/(2n+1)$ ?

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Answer 1 · 2014-10-13T21:32:46

First expand the series for each value of n $\frac{n}{2 n + 1}$ $n/(2n+1)$ = $\frac{1}{2 (1) + 1}$ $1/(2(1)+1$ + $\frac{2}{2 (2) + 1}$ $2/(2(2)+1$ + $\frac{3}{2 (3) + 1}$ $3/(2(3)+1$ + $\frac{4}{2 (4) + 1}$ $4/(2(4)+1$ + $\frac{5}{2 (5) + 1}$ $5/(2(5)+1$

Next, perform the operations in the denominator...

$\frac{1}{3}$ $1/(3)$ + $\frac{2}{5}$ $2/5$ + $\frac{3}{7}$ $3/7$ + $\frac{4}{9}$ $4/9$ + $\frac{5}{11}$ $5/11$

Now, to add fractions we need a common denominator... in this case it's $3465$ $3465$

Next, we have to multiply each numerator and denominator by the missing components...

$\frac{1}{3}$ $1/3$ gets multiplied by $1155$ $1155$ giving $\frac{1155}{3465}$ $1155/3465$

(Divide the $3465$ $3465$ by $3$ $3$ to get $1155$ $1155$ and divide the rest by the given denominator.)

$\frac{2}{5} \cdot \frac{693}{693} = \frac{1386}{3465}, \frac{3}{7} \cdot \frac{495}{495} = \frac{1485}{3465}, \frac{4}{9} \cdot \frac{385}{385} = \frac{1540}{3465} and \frac{5}{11} \cdot \frac{315}{315} = \frac{1575}{3465}$ $2/5*693/693=1386/3465, 3/7*495/495=1485/3465, 4/9*385/385=1540/3465 and 5/11*315/315=1575/3465$

Now simply add the numerators together... $\frac{1155 + 1386 + 1485 + 1540 + 1575}{3465}$ $(1155+1386+1485+1540+1575)/3465$

giving $\frac{7141}{3465}$ $7141/3465$ .

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How do you evaluate the sum represented by $5 \sum n = 1 \frac{n}{2 n + 1}$ $sum_(n=1)^5n/(2n+1)$ ?

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How do you evaluate the sum represented by 5∑n=1n2n+1sum_(n=1)^5n/(2n+1) ?

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How do you evaluate the sum represented by $5 \sum n = 1 \frac{n}{2 n + 1}$ $sum_(n=1)^5n/(2n+1)$ ?