A Sum with Fractions and Roots

Since this sum looks pretty complex, let's try to simplify it down.

Let's look at the first fraction, $\frac{1}{\sqrt{1}+\sqrt{2}}$ . If we multiply this by $\frac{\sqrt{1}-\sqrt{2}}{\sqrt{1}-\sqrt{2}}$ , we obtain:

$\frac{\sqrt{1}-\sqrt{2}}{(\sqrt{1}+\sqrt{2})\times(\sqrt{1}-\sqrt{2})}$

$=\frac{\sqrt{1}-\sqrt{2}}{1+\sqrt{2}-\sqrt{2}-2}$

$=\frac{\sqrt{1}-\sqrt{2}}{-1}$

$=\sqrt{2}-\sqrt{1}$ .

Isn't that neat? We can try a couple more fractions, and soon realize that all of them are the same.

$\frac{1}{\sqrt{a}+\sqrt{a+1}}=\sqrt{a+1}-\sqrt{a}$

This is because the denominator will always be -1. (The two middle terms of the denominator when we are multiplying by the conjugate cancel out and $a-(a+1)=-1$ )

Now, we have simplified our complex sum to something much simpler.

$\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\ldots-\sqrt{98}+\sqrt{100}-\sqrt{99}$

All the terms besides the first and last one cancel out, so we are left with

$=\sqrt{100}-\sqrt{1}$

$=10-1=\boxed{9}$

The sum telescopes as follows: $\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}=\sqrt{n+1}-\sqrt{n}$ Then, the sum is $(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\cdots+(\sqrt{100}-\sqrt{99})=\sqrt{100}-\sqrt{1}=\boxed{9}$

Each term of the sum is in the form $\dfrac{1}{\sqrt{k} + \sqrt{k+1}}$ for some positive integer $k.$ Rationalizing this expression, we have

$\begin{aligned} \dfrac{1}{\sqrt{k} + \sqrt{k+1}} &= \dfrac{\sqrt{k+1} - \sqrt{k}}{(\sqrt{k} + \sqrt{k+1})(\sqrt{k+1} - \sqrt{k})} \\ &= \dfrac{\sqrt{k+1}-\sqrt{k}}{1} = \sqrt{k+1}-\sqrt{k}. \end{aligned}$

Therefore, the given sum is equal to

$(\sqrt2-\sqrt1)+(\sqrt3-\sqrt2)+\ldots+(\sqrt{100}-\sqrt{99})$

and most of the terms cancel to give $-\sqrt1 + \sqrt{100} = \boxed{9}.$

this can be rewritten as :

$\sum_{n=1}^{99}\frac{1}{\sqrt{n} + \sqrt{n+1}}=$

$\sum_{n=1}^{99}\frac{\sqrt{n} - \sqrt{n+1}}{(\sqrt{n} + \sqrt{n+1})(\sqrt{n} - \sqrt{n+1})} =$

$\sum_{n=1}^{99}\frac{\sqrt{n} - \sqrt{n+1}}{(n)-(n+1)} =$

$\sum_{n=1}^{99}-(\sqrt{n} - \sqrt{n+1}) = \sum_{n=1}^{99} (\sqrt{n+1} - \sqrt{n})$

so it's $(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{100} - \sqrt{99}) = 10 - 1 = 9$

Using conjugation, the sum is $S=\frac{1}{\sqrt1+\sqrt2}\times\frac{\sqrt1-\sqrt2}{\sqrt1-\sqrt2}+\frac{1}{\sqrt2+\sqrt3}\times\frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}\cdots+\frac{1}{\sqrt99+\sqrt100}\times\frac{\sqrt99-\sqrt100}{\sqrt99-\sqrt100}$ .

Simplifying, $S=\frac{\sqrt1-\sqrt2}{(\sqrt1+\sqrt2)(\sqrt1-\sqrt2)}+\frac{\sqrt2-\sqrt3}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)}+\cdots+\frac{\sqrt99-\sqrt100}{(\sqrt99+\sqrt100)(\sqrt99-\sqrt100)}$ .

Rationalizing the denominator with $(a+b)(a-b)$ , $S=\frac{\sqrt1-\sqrt2}{-1}+\frac{\sqrt2-\sqrt3}{-1}+\cdots+\frac{\sqrt99-\sqrt100}{-1}$ .

Adding the fractions, $S=\frac{\sqrt1-\sqrt2+\sqrt2-\sqrt3+\sqrt3-\sqrt4+\cdots+\sqrt99-\sqrt100}{-1}$ .

Combining like terms, $S=\frac{\sqrt1-\sqrt100}{-1}=\frac{1-10}{-1}=\frac{-9}{-1}=9$ .

Rationalize every term, you will get- √100-√1= 9

This question can easily be solved using the Telescoping series :

Series can be written as $\Sigma \frac{1}{(\sqrt{n})+(\sqrt{n+1})}$ which we can rationalize and write as $\Sigma \sqrt{n+1}-\sqrt{n}$ (upper and lower limits are 99 and 1 respectively).Putting the limits $\sqrt{10}-\sqrt1 = \boxed9$

Rationalizing yields: sqrt(2)-sqrt(1)-sqrt(2)+sqrt(3)......-sqrt(99)+sqrt(100). All terms except sqrt(100) and -sqrt(1) cancel out. Therefore, the expression equals sqrt(100)-sqrt(1)=10-1=9

All terms are of the form $\frac{1}{(\sqrt{n}+\sqrt{n+1}}$ . When we multiply the numerator and the denominator by the conjugate of the denominator, we get $\frac{\sqrt{n}-\sqrt{n+1}}{-1}$ , which is equal to $\sqrt{n+1}-\sqrt{n}$ . Our problem now looks like this: $\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}$ . We can see that this is a telescoping sum. After simplifying with mass cancellation, we are left with $\sqrt{100}-\sqrt{1}=10-1=\boxed{9}$ .

Rationalize the terms so 1/(sqrt(1)+sqrt(2)) becomes (sqrt(2)-sqrt(1)) similarly 1/(sqrt(2)+sqrt(3)) becomes (sqrt(3)-sqrt(2)) and 1/(sqrt(3)+sqrt(4)) becomes (sqrt(4)-sqrt(3))
and so on
last term is 1/(sqrt(99)+sqrt(100)) becomes (sqrt(100)-sqrt(99)) adding all the terms we will get sqrt(100)-sqrt(1)=10-1=9

rationalise each and every term and you wiil get '1' as denominator in each term ...every term will be cancelled except sqtr(100)-sqrt(1) ....hence answer will be 9.

Let's start by writing it as a summation with sigma notation: $\sum_{n=1}^{99} \frac{1}{\sqrt{n}+\sqrt{n+1}}$

Now, we can rationalize the denominator: $\sum_{n=1}^{99} \frac{1}{\sqrt{n}+\sqrt{n+1}} \cdot \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}$

And simplify: $\begin{aligned} \sum_{n=1}^{99} \frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1})^2-(\sqrt{n})^2} &= \sum_{n=1}^{99} \frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-(n)} \\ &= \sum_{n=1}^{99} \frac{\sqrt{n+1}-\sqrt{n}}{1} \\ &= \sum_{n=1}^{99} \left( \sqrt{n+1}-\sqrt{n}\right) \end{aligned}$

Because of the commutative law of addition, we have a rule that $\sum_{n=1}^{n=k} (a_n \pm b_n) = \sum_{n=1}^{n=k}a_n \pm \sum_{n=1}^{n=k}b_n .$

Following that rule gets us $\sum_{n=1}^{99} \left( \sqrt{n+1}-\sqrt{n}\right) = \sum_{n=1}^{99} \sqrt{n+1}- \sum_{n=1}^{99}\sqrt{n} .$

Now, we can make the substitution $m=n+1$ in the first summation. Doing so makes the lower limit of $n=1$ become $m=2$ and the upper limit of $99$ become $100$ , giving us $\sum_{m=2}^{100} \sqrt{m}- \sum_{n=1}^{99}\sqrt{n} .$

We can rewrite these sums to have the same limits by removing a term from each of them: $\left(\sum_{m=2}^{99} \sqrt{m} +\sqrt{100}\right) - \left( \sum_{n=2}^{99}\sqrt{n} + \sqrt{1} \right)$

Distributing the negative and rearranging the terms gives us the following: $\begin{aligned} \sum_{m=2}^{99} \sqrt{m} +\sqrt{100} - \sum_{n=2}^{99}\sqrt{n} - \sqrt{1} &= \sum_{m=2}^{99} \sqrt{m} - \sum_{n=2}^{99}\sqrt{n} +\sqrt{100} - \sqrt{1} \\&= \sum_{m=2}^{99} \sqrt{m} - \sum_{n=2}^{99}\sqrt{n} +10-1 \\&= \sum_{m=2}^{99} \sqrt{m} - \sum_{n=2}^{99}\sqrt{n} +9 \end{aligned}$

Even though the two summations use different index variables, they are the exact same thing. Using this fact, we get $\left( \sum_{m=2}^{99} \left( \sqrt{m}\right) - \sum_{n=2}^{99}\sqrt{n} \right) +9 = 0+9=\boxed{9}$

Can be -11 as well.

its easy understanding, rationalise the denominator of every term and then simplify

Factorise each term in the series. Subsequent terms will keep cancelling, leaving only sqrt(100)-sqrt(1)

$\frac{1}{\sqrt{x}+\sqrt{x+1}}=\frac{1}{\sqrt{x}+\sqrt{x+1}}*\frac{\sqrt{x}-\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}=\frac{\sqrt{x}-\sqrt{x+1}}{-1}=\sqrt{x+1}-\sqrt{x}$

so the problem now could be written as follow

$(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+..............+(\sqrt{100}-\sqrt{99})=\sqrt{100}-\sqrt{1}=\boxed{9}$

T(r) = 1/(sqrt{r} + sqrt{r+1}) = sqrt{r+1} - sqrt{r} ...............{After rationalisation}

Therefore, from ques. T(1) + T(2) +...+ T(99) = sqrt{100} - sqrt{1} = 10 - 1 = 9. {Ans.}