Find the sum of left(1-frac{1}{n+1}right)+lef | vedclass

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(B) The given expression is $\sum_{k=1}^{n} \left(1-\frac{k}{n+1}\right)$.This can be expanded as $\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\

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$\frac{1}{2} n$

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(B) The given expression is $\sum_{k=1}^{n} \left(1-\frac{k}{n+1}\right)$.This can be expanded as $\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\right)+\cdots+\left(1-\frac{n}{n+1}\right)$.There are $n$ terms in this series,so we can write it as $n - \left(\frac{1}{n+1} + \frac{2}{n+1} + \cdots + \frac{n}{n+1}\right)$.Factoring out the denominator,we get $n - \frac{1}{n+1} (1+2+3+\cdots+n)$.Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we substitute this into the expression:$n - \frac{1}{n+1} \cdot \frac{n(n+1)}{2}$.Canceling $(n+1)$ from the numerator and denominator,we get $n - \frac{n}{2}$.Thus,the sum is $\frac{n}{2}$.

Find the sum of $\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\right)+\left(1-\frac{3}{n+1}\right)+\cdots+\left(1-\frac{n}{n+1}\right)$

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