Question

Please give the answer of the question attached with explanation​

Answer 1

Required Answer:-

Given to find:

The value of $\rm \bigg[ \bigg(1 - \dfrac{1}{n + 1} \bigg) + ... \bigg(1 - \dfrac{n}{n + 1} \bigg) \bigg]$

Solution:

We have,

$\rm \bigg[ \bigg(1 - \dfrac{1}{n + 1} \bigg) + \bigg(1 - \dfrac{2}{n + 1} \bigg) + ... \bigg(1 - \dfrac{n}{n + 1} \bigg) \bigg]$

This can be written as,

$\rm = \bigg[(1 + 1 + ..n) - \bigg( \dfrac{1}{n + 1} + .. \dfrac{n}{n + 1} \bigg)\bigg]$

Sum of n 1s is n only. Therefore,

$\rm = \bigg[n- \bigg( \dfrac{1}{n + 1} \times (1 + 2 + ..n)\bigg)\bigg]$

Now, you know that, sum of first n natural numbers is n(n + 1)/2. Putting the value, we get,

$\rm = \bigg[n- \bigg( \dfrac{1}{n + 1} \times \dfrac{n(n + 1)}{2} \bigg)\bigg]$

Now, n + 1 in both Numerator and Denominator gets cancelled out. So,

$\rm = \bigg[n- \dfrac{n}{2} \bigg]$

$\rm = \dfrac{n}{2}$

Hence, the sum of the series is equal to n/2. So, option (A) is our required answer.

Answer:

• Option (A) is the correct answer for this question.

Answer 2

ans is A (n/2) let us c how

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