Radical Fractions, Man!

Let us first write the sum in summation notation, after we receive a little inkling of what should come next.

$\sum_{k=1}^{9999} \frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}$

Let us closely examine the fraction $\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}$ . If we multiply both the numerator and the denominator by $k\sqrt{k+1}-(k+1)\sqrt{k}$ , we obtain

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

Isn't that beautiful? Our sum now decomposes into two very simple quantities.

$\begin{aligned} \sum_{k=1}^{9999} \frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}} = \sum_{k=1}^{9999} \frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1} \\ = \sum_{k=1}^{9999} \frac{\sqrt{k}}{k} - \sum_{k=1}^{9999} \frac{\sqrt{k+1}}{k+1} \\ = (\frac{\sqrt{1}}{1}+\frac{\sqrt{2}}{2} \ldots \frac{\sqrt{9999}}{9999}) - (\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}...\frac{\sqrt{10000}}{10000}) \\ = \frac{\sqrt{1}}{1}-\frac{\sqrt{10000}}{10000} \\ = \frac{10000-100}{1000} \\ = \frac{99}{100} \end{aligned}$

Our desired sum is $\boxed{199}$ .

See that, we will simplify the fraction, that is :

$\sum_{i= 1}^{9999} \frac{1}{a\sqrt{a+1} + (a+1)\sqrt{a}} = \sum_{i= 1}^{9999} \frac{1}{a\sqrt{a+1} + (a+1)\sqrt{a}} \times \frac{(a+1)\sqrt{a} - a\sqrt{a+1}}{(a+1)\sqrt{a} - a\sqrt{a+1}}$ . Implies :

$= \sum_{i= 1}^{9999} \frac{(a+1)\sqrt{a} - a\sqrt{a+1}}{a(a+1)} = \sum_{i= 1}^{9999} ( \frac{(a+1)\sqrt{a}}{a(a+1)} - \frac{a\sqrt{a+1}}{a(a+1)} )$

$= \sum_{i= 1}^{9999} ( \frac{1}{\sqrt{a}} - \frac{1}{\sqrt{a+1}} ) = \frac{1}{1} - \frac{1}{\sqrt{10000}} = 1 - \frac{1}{100} = \frac{99}{100}$

$\implies a = 99$ , and $b = 100$ . So,

$a + b = 99 + 100 = \boxed{199}$

Let us consider a general term of the following series -

$\frac{1}{(a+1)(\sqrt{a}) + (a)(\sqrt{a+1})}$

This can be rationalized to get

$\\ \frac{1}{(a+1)(\sqrt{a}) + (a)(\sqrt{a+1})} \\$ $\\ =\frac{1}{(a+1)(\sqrt{a}) + (a)(\sqrt{a+1})} \ \times \frac{(a+1)(\sqrt{a}) -((a)(\sqrt{a+1})}{(a+1)(\sqrt{a}) - (a)(\sqrt{a+1})} \\$ $\\ = \frac{(a+1)(\sqrt{a}) - (a)(\sqrt{a+1})}{(a)(a+1)^2-(a)^2(a+1)} \\$ $\\ = \frac{(a+1)(\sqrt{a}) - (a)(\sqrt{a+1})}{(a)(a+1)[ \ (a+1)-(a) \ ]} \\$ $\\ = \frac{(a+1)(\sqrt{a}) - (a)(\sqrt{a+1})}{(a)(a+1)} \\$ $\\ = \frac{(a+1)(\sqrt{a})}{(a)(a+1)} - \frac{(a)(\sqrt{a+1})}{(a)(a+1)} \\$ $\\ = \frac{1}{\sqrt{a}} - \frac{1}{\sqrt{a+1}} \\$

Now the given series can be written as -

$\bigg(\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}}\bigg) + \bigg(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\bigg) + \bigg(\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}}\bigg) + \ \dots \ +$

$\bigg(\frac{1}{\sqrt{9998}} - \frac{1}{\sqrt{9999}}\bigg) + \bigg(\frac{1}{\sqrt{9999}} - \frac{1}{\sqrt{10000}}\bigg)\\$

which, on cancellation reduces to

$1 - \frac{1}{100} = \frac{99}{100}$

So the answer is $99 + 100 = 199$

$\frac{1}{( a\sqrt{b}+b\sqrt{a})}$ = $\frac{1\times ( a\sqrt{b}-b\sqrt{a})}{( a\sqrt{b}+b\sqrt{a}) \times (a\sqrt{b}-b\sqrt{a})}$ = $\frac{ ( a\sqrt{b}-b\sqrt{a})}{ a^2b-b^2a}$ = $\frac{ ( a\sqrt{b}-b\sqrt{a})}{ ab(a-b)}$

in the above series a-b= -1

so,

$\frac{ ( a\sqrt{b}-b\sqrt{a})}{ ab(a-b)}$ = $\frac{ ( a\sqrt{b}-b\sqrt{a})}{ - ab}$ = $\frac{ -1}{\sqrt{b}}+\frac{ 1}{\sqrt{a}}$ and

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

$\frac{1}{( 2\sqrt{3}+3\sqrt{2})}$ = $\frac{ -1}{\sqrt{3}}+\frac{ 1}{\sqrt{2}}$

..

..

..

$\frac{1}{( 9999\sqrt{10000}+10000\sqrt{9999})}$ = $\frac{ -1}{\sqrt{10000}}+\frac{ 1}{\sqrt{9999}}$

adding all the terms we get

$\frac{ -1}{\sqrt{2}}+\frac{ 1}{\sqrt{1}}$ + $\frac{ -1}{\sqrt{3}}+\frac{ 1}{\sqrt{2}}$ +..........+ $\frac{ -1}{\sqrt{10000}}+\frac{ 1}{\sqrt{9999}}$

= 1- $\frac{1}{\sqrt{10000}}$ = 1- $\frac{1}{100}$ = $\frac{99}{100}$

Note that

$\frac{1}{k\sqrt{k+1}+ (k+1)\sqrt{k}}$

= $-\frac{k\sqrt{k+1}-(k+1)\sqrt{k}}{k(k+1)}$

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

Therefore, the desired sum is equal to $1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{9999}}-\frac{1}{\sqrt{10000}}=1-\frac{1}{\sqrt{10000}}=\frac{99}{100}$

Thus, we have $a+b=99+100=199$

$\frac{1}{i\sqrt{i+1}+(i+1)\sqrt{i}} = \frac{(i+1)\sqrt{i}-i\sqrt(i+1)}{i(i+1)} = \frac{\sqrt{i}}{i} -\frac{\sqrt{i+1}}{i+1}$ . Thus the given sum is $\frac{\sqrt{1}}{1} -\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} -\frac{\sqrt{3}}{3}+\cdots + \frac{\sqrt{9998}}{9998} -$ $\frac{\sqrt{9999}}{9999}+\frac{\sqrt{9999}}{9999} -\frac{\sqrt{10000}}{10000} = 1-.01 = \frac{99}{100}$ , so the answer is $99+100=\fbox{199}$ .

We first find a generic simplification of each added term. We can write each term in the form $\frac{1} {a\sqrt{a+1} +(a+1)\sqrt{a}}$ = $\frac{1} {a\sqrt{a+1} +(a+1)\sqrt{a}} \cdot \frac{a\sqrt{a+1} -(a+1)\sqrt{a}} {a\sqrt{a+1} -(a+1)\sqrt{a}}$ = $\frac{a\sqrt{a+1} -(a+1)\sqrt{a}} {a^2(a+1) -(a+1)^2a}$ . Simplifying, $\frac{(a+1)\sqrt{a} -a\sqrt{a+1}} {a(a+10}$ . We know separate this fraction into two fractions, getting us $\frac{(a+1)\sqrt{a}} {a(a+1)} - \frac{a\sqrt{a+1}} {a(a+1)}$ =
$\frac{1} {\sqrt{a}} - \frac{1} {\sqrt{a+1}}$ . Thus we find that the sequence telescopes, with terms cancelling out. Now, we can write our series as, with cancelled terms $\frac{1} {\sqrt{1}} - \frac{1} {\sqrt{10000}}$ , which is equal to 99/100. Thus our answer is 99+100=199.

1/[n\sqrt{n+1}+ (n+1)\sqrt{n}] = [(n+1)\sqrt{n} - n\sqrt{n+1}] / [(n+1)\sqrt{n} - n\sqrt{n+1}][n\sqrt{n+1}+ (n+1)\sqrt{n}] = [(n+1)\sqrt{n} - n\sqrt{n+1}] / [(n+1)^2 (n) -n^2 (n+1)] = [(n+1)\sqrt{n} - n\sqrt{n+1}] / [n(n+1)(n+1-n)] = [(n+1)\sqrt{n}] / [n(n+1)] - [n\sqrt{n+1}] / [n(n+1)] = 1/sqrt{n} - 1/sqrt{n+1}

Hence the telescoping sum is actually 1/sqrt{1} - 1/sqrt{2} + 1/sqrt{2}- 1/sqrt{3}+ ...+1/sqrt{9998}- 1/sqrt{9999}+ 1/sqrt{9999} - 1/sqrt{10000} = 1/1 - 1/100 = 99/100.

Therefore a= 99 and b=100. The result follows that a+b = 199.

Let $S=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...+\frac{1}{9999\sqrt{10000}+10000\sqrt{9999}}$

$S=\displaystyle\sum_{n=1}^{9999} \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}$

Rationalizing the denominator,

$S=\displaystyle\sum_{n=1}^{9999} \frac{n\sqrt{n+1}-(n+1)\sqrt{n}}{n^2(n+1)-(n+1)^2(n)}$

$S=\displaystyle\sum_{n=1}^{9999} \frac{n\sqrt{n+1}-(n+1)\sqrt{n}}{-n(n+1)}$

$S=\displaystyle\sum_{n=1}^{9999} -\frac{\sqrt{n+1}}{n+1}+\frac{\sqrt{n}}{n}$

Expanding,

$S=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{9999}}{9999}-\frac{\sqrt{10000}}{10000}$

Simplifying,

$S=\frac{\sqrt{1}}{1}-\frac{\sqrt{10000}}{10000}=1-\frac{100}{10000}=\frac{99}{100}$

Therefore, $a+b=199$ .

Note that each expression is equivalent to 1/(nsqrt(n+1)+(n+1)sqrt(n)), for 1<=n<=9999. 1/(nsqrt(n+1)+(n+1)sqrt(n)) * ((n+1)sqrt(n)-nsqrt(n+1))/((n+1)sqrt(n)-nsqrt(n+1)) =((n+1)sqrt(n)-nsqrt(n+1))/(n(n+1)^2-n^2(n+1)) =((n+1)sqrt(n)-nsqrt(n+1))/(n^2+n) =((n+1)sqrt(n)-nsqrt(n+1))/(n(n+1)) =1/(sqrtn)-1/(sqrt(n+1)). This telescopes, so it is equivalent to 1-1/sqrt(10000)=1-100=99/100. Therefore, the answer is 199.

To get rid of the square roots in the denominator, we use $x^2 - y^2 = (x+y)(x-y)$ . Therefore, the first term will become $\frac {1} {1\sqrt {2} + 2\sqrt {1}} = \frac {1* (2\sqrt {1} - 1\sqrt {2}) } { (2\sqrt {1})^2 - (1\sqrt {2})^2} = \frac {1* (2\sqrt {1} - 1\sqrt {2}) } {4 - 2} = \frac {1* (2\sqrt {1} - 1\sqrt {2}) } {1 * 2} = \frac { \sqrt {1}}{1} - \frac {\sqrt {2}}{2}$ .

Using the same steps, the second term will be $\frac { \sqrt {2}}{2} - \frac {\sqrt {3}}{3}$ and the third term will be $\frac { \sqrt {3}}{3} - \frac {\sqrt {4}}{4}$ and so on......

Then, by cancelling out the positive and negative fractions which are the same, the expression will be :

$\frac { \sqrt {1}} {1} - \frac {\sqrt {2}} {2} + \frac {\sqrt {2}} {2} - \frac {\sqrt {3}} {3} + \frac { \sqrt {3}} {3} - \frac {\sqrt {4}} {4} ... + \frac { \sqrt {9999}} {9999} - \frac {\sqrt {10000}} {10000}$

= $\frac { \sqrt {1}} {1} - \frac {\sqrt {10000}} {10000}$

= $1 - \frac {1} {100} = \frac {99} {100}$ .

So, $a = 99 , b = 100, a + b = 199.$

Let $n=9999$ . The given expression is

$\displaystyle \sum_{r=1}^{n} \frac{1}{r\sqrt{r+1}+(r+1)\sqrt{r}} = \sum_{r=1}^{n} \frac{r\sqrt{r+1}-(r+1)\sqrt{r} }{r^2(r+1)-(r+1)^2r}$

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

Writing down a few terms,

$\displaystyle \sum_{r=1}^{n} \frac{1}{\sqrt{r}}-\frac{1}{\sqrt{r+1}}=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-........-\frac{1}{\sqrt{10^4}}$

We notice that most of the terms cancel and we are left with

$\displaystyle 1-\frac{1}{\sqrt{10^4}}=1-\frac{1}{100}=\frac{99}{100}$

Hence, the answer is $99+100=\fbox{100}$ .

Nth term =

1

(n sqrt(n+1)+(n+1) sqrt(n))

(n+1)*sqrt(n)-n*sqrt(n+1)

= -------------------------------------------------------------------------
(n sqrt(n+1)+(n+1) sqrt(n))* ((n+1) sqrt(n)-n sqrt(n+1))

(n+1) sqrt(n)-n sqrt(n+1))
= -----------------------------------------
((n+1)^2) n-(n^2) (n+1)

(n+1)*sqrt(n)-n*sqrt(n+1)

= ----------------------------------------------
(n^3)+2*(n^2)+n –(n^3) –(n^2)

(n+1) sqrt(n)-n sqrt(n+1) = ---------------------------------------- n*(n+1)

= 1/sqrt(n) – 1/sqrt(n+1)

So, 1st term = 1/1 – 1/sqrt(2) 2nd term = 1/sqrt(2) - 1/sqrt(3)

……………………………………………….. and so on 9999th term = 1/sqrt(9999) – 1/sqrt(10000)

Adding we get Total Sum of the series = 1- 1/sqrt(10000) = 1-1/100=99/100 So the answer is 99+100, i.e. 199.

We claim that $\sum_{i=1}^{k^{2}-1}\frac{1}{i\sqrt{i+1}+(i+1)\sqrt{i}}=1-\frac{1}{k}, \forall k\in\mathbb N$ So the aum above is $\frac{99}{100}.$

Base case: $i=1$, so that $\frac{1}{2+\sqrt{2}}=\frac{\frac{1}{1+\sqrt{2}}}{\frac{2+\sqrt{2}}{1+\sqrt{2}}}=\frac{\sqrt{2}-1}{\sqrt{2}}=1-\frac{1}{\sqrt{2}}$

Inductive step. Let the problem to be true for $i=n-1,$ so that $\sum_{i=1}^{n-1}\frac{1}{i\sqrt{i+1}+(i+1)\sqrt{i}}=1-\frac{1}{\sqrt {n}}.$

For $i=n+1$ it is sufficient to prove that $1-\frac{1}{\sqrt{n}}+\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}=1-{1}{\sqrt{n+1}},$ or \frac{1}{\sqrt{n}}-{1}{\sqrt{n+1}}=\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}.$ But $\frac{1}{\sqrt{n}}-{1}{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}$ and $(\sqrt{n+1}-\sqrt{n})(n\sqrt{n+1}+(n+1)\sqrt{n})=n(n+1)+(n+1)\sqrt{n(n+1)}-n\sqrt{n(n+1)}-n(n+1)=\sqrt{n(n+1)},$ so \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}.$ This completes the proof.

The expression can be expressed as $\displaystyle \sum_{i=1}^{99} \frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}$ . Multiply the denominator and numerator by $(n+1) \sqrt{n}-n \sqrt{n+1}$ and we will get $\displaystyle \sum_{i=1}^{99} \frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{n(n+1)} = \displaystyle \sum_{i=1}^{99} \frac{ \sqrt{n}}{n}- \frac{ \sqrt{n+1}}{n+1}$ (We get the denominator using difference of squares). Since this is a telescoping sum, the intermediate terms cancel and we get $1- \frac{1}{100} = \frac{99}{100}$ .

Whenever you see a irrational denominator try to rationalize it (most commonly done by multiplying by the conjugate of the denominator). By rationalizing the first fraction we get [2-sqrt(2)]/2, then if we add the second term and rationalize the answer we obtain [3-sqrt(3)]/3. Thus we notice a pattern emerging. The sum of (n-1)th terms is given by [n-sqrt(n)]/n. Therefore the last term is 10000-sqrt(10000) all divided by 10000 = 99/100. Therefore a = 99 and b = 100 (satisfying the condition that a and b are coprime positive integers) therefore a+b = 99+100 = 199.

Here, T(n) = 1/[n.root(n+1) + (n+1).rootn] = 1/rootn - 1/root(n+1). (Using key techniques) by telescopic calculation we get s(n) = 1-1/100 = 99/100 a=99, b= 100 a + b = 99 + 100 = 199, which is the solution

1/(1√2+2√1) Can be simplified as (2√1−1√2)/(4−2)=√1−1/√2. similarly 1/(2√3+3√2) can be simplified as (3√2−2√3)/(18−12)=1/√2−1√3 Hence we observe that its a repetition in which each cluster(eg 1/√2) occurs twice in alternating sign in the summation, except the first number in the sequence and the last number the last number is -9999√10000/(10000∗10000∗9999 −9999∗9999∗10000) = −1/√10000= −1/100, Therefore by telescopic summation the final answer is 1−1/100= 99/100, in which 99 and 100 are co-prime so a=99 and b=100, Hence a+b=199 .

Tools needed to solve this problem:

• Telescoping Method

Main Idea :

1. Determine the General Form .

2. Analyse it and break it to the form: $f(n)-f(n+1)$

3. Then we can obtain something like: $f(1)-f(2)+f(2)-f(3)+f(3)-f(4)+...+f(n)-f(n+1)$

4. The answer to the problem will be

$f(1)-f(n+1)$

---

Solution

Determine the General Form

By observing, the General Form is

$\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}$

Analyse

Now, we will to remove the square roots in the denominator

$\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}$

$=\frac{n\sqrt{n+1}-(n+1)\sqrt{n}}{n^2(n+1)-(n+1)^2n}$

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

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

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

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

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

Or you may prefer

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

Thus,

$\displaystyle \sum_{i=1}^{9999} \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$

$=1-\frac{1}{\sqrt10000}$

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

$=\frac{99}{100}$

Hence, $a+b=199$

We observe that each term is of the form

$\frac{1}{n \sqrt{n+1} + (n+1) \sqrt{n} }$ . We need to simplify this.

Rationalizing and simplifying

$\frac{ (n+1) \sqrt{n} - n \sqrt{n+1} }{n^2 +n}$ $= \frac{ (n+1) \sqrt{n} - n \sqrt{n+1} }{n(n+1)}$ $= \frac{1}{ \sqrt{n} } - \frac{1}{ \sqrt{n+1} }$

Thus we get a simplified version. So the given expression becomes

$[ \frac{1}{ \sqrt{1} } - \frac{1}{ \sqrt{2} } ] + [ \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{3} } ] +.......+ [ \frac{1}{ \sqrt{9999} } - \frac{1}{ \sqrt{10000} } ]$

All the terms except first and the last is canceled out as negative fraction of one term cancel out the positive term of other. So above expression is equal to

$\frac{1}{ \sqrt{1} } - \frac{1}{ \sqrt{10000} }$

$= 1 - \frac{1}{100} = \frac{99}{100}$

Thus we get our requires answer as $99 + 100 = \boxed{199}$

Note that $\begin{aligned} \frac{1}{i\sqrt{i + 1} + (i + 1)\sqrt{i}} &= \frac{(i + 1)\sqrt{i} - i\sqrt{i + 1}}{i(i + 1)} \\ &= \frac{\sqrt{i + 1} - \sqrt{i}}{\sqrt{i}\sqrt{i + 1}} \\ &= \frac{1}{\sqrt{i}} - \frac{1}{\sqrt{i + 1}} \end{aligned}$ Hence $\begin{aligned}\sum_{i = 1}^{9999} \frac{1}{i\sqrt{i + 1} + (i + 1)\sqrt{i}} &= \sum_{i = 1}^{9999} \left( \frac{1}{\sqrt{i}} - \frac{1}{\sqrt{i + 1}} \right ) \\ &= \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{10000}} \\ &= \frac{99}{100} \end{aligned}$ Thus $a + b = \boxed{199}$

$\displaystyle \sum_{r = 1}^{9999} \frac{1}{r\sqrt{r + 1} + (r + 1)\sqrt{r}}$

$=\displaystyle \sum_{r = 1}^{9999} \frac{(r + 1)\sqrt{r} - r\sqrt{r + 1}}{(r +1 )^2r - r^2(r + 1)}$

$=\displaystyle \sum_{r=1}^{9999} \frac{(r + 1)\sqrt{r} - r\sqrt{r +1}}{r(r+1)}$

$=\displaystyle \sum_{r =1 }^{9999} \bigg(\frac{1}{\sqrt{r}} - \frac{1}{\sqrt{r + 1}}\bigg)$

$= \frac{1}{1} - \frac{1}{\sqrt{9999 + 1}}$

$= \boxed{\frac{99}{100}}$

$\frac{1}{(n \times \sqrt{n+1}) + ((n+1) \times \sqrt{n}) }$

$= \frac{1}{ \sqrt{n} \times \sqrt{n+1} \times (\sqrt{n+1} + \sqrt{n}) }$

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

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

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

By using the method of Telescoping Series , we get

$\frac{1}{1} - \frac{1}{ \sqrt{10000} } = 1 - \frac{1}{100} = \frac{99}{100}$

Hence we get, $a = 99, b = 100$ Therefore $a + b = \boxed{199}$

By rationalizing the fractions, we find that the expression actually telescopes.

$\frac {1}{1 \sqrt{2} + 2 \sqrt{1}} + \frac {1}{2 \sqrt{3} + 3 \sqrt{2}} + ... + \frac {1}{9999 \sqrt{10000} + 10000 \sqrt{9999}}$

$=\frac {2 \sqrt {1} - 1\sqrt {2}}{1 \times 2} + \frac {3 \sqrt {2} - 2\sqrt {3}}{2 \times 3} + ... +\frac {10000 \sqrt {9999} - 9999\sqrt {10000}}{9999 \times 10000}$

$=\frac {\sqrt{1}}{1} - \frac {\sqrt{2}}{2} + \frac {\sqrt{2}}{2} - \frac {\sqrt{3}}{3} + ... +\frac {\sqrt{9999}}{9999} - \frac {\sqrt{10000}}{10000}$

$=1 - \frac {\sqrt{10000}}{10000}$

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

$=\frac {99}{100}$

Thus, $\frac {a}{b} = \frac {99}{100}$ , and $a + b = 100 + 99 = \boxed {199}$

We can rewrite 1/{a\sqrt{a+1}+(a+1)\sqrt{a}} as 1/\sqrt{a}-1/\sqrt{a+1} so it telescopes and we get 1/1-1/100=99/100 for answer of 199

The expression $\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}$ is too ugly. So, we use the identity $a^2-b^2=(a+b)(a-b)$ to simplify the expression.

Note that $\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}=\frac{k\sqrt{k+1}-(k+1)\sqrt{k}}{k^2(k+1)-k(k+1)^2}=\frac{k\sqrt{k+1}-(k+1)\sqrt{k}}{k(k+1)}=\frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$ .

Then the sum telescopes perfectly to $1-\frac{1}{\sqrt{10000}}=\frac{99}{100}$ .Thus, $a+b=199$ .

rationalise the denominator and factorise the expression for the nth term you will get 1/(n)^1/2-1/(n+1)^1/2. using telescopic summation you shall get 1-1/100=99/100, a+b=99+100=199

It's pretty simple. All you need to do is reduce the series into a telescopic one, i.e., one with alternating positive and negative signs. First, recognize the general rth term, which is (r sqrt(r+1) + sqrt(r) (r+1))^-1; then what you gotta do is manipulate a little. Taking common factors, you get, [ sqrt(r) sqrt(r+1) { sqrt(r) + sqrt(r+1) } ]^-1. Multiply numerator and denominator by sqrt(r) + sqrt(r+1) to get rid of the radical, and split into two fractions. You'll end up with Tr = [(sqrt(r))^-1] - [sqrt(r+1))^-1]. On summing up till n terms, except for first and last term, all the other terms get cancelled out. Therefore, sum is 1 - [sqrt(n+1))^-1]. Substituting n as 10000 from the above problem, you get the required answer.

Looking at the terms of the sum, we notice that all the terms are of the form $\frac {1}{a{\sqrt{a+1}} + (a+1)\sqrt{a}}$ , where $a=1,2,3,\dots, 9999$ .

By rationalizing the fraction and manipulating it, we eventually get it into the form:

$\frac {1}{a{\sqrt{a+1}} + (a+1)\sqrt{a}}$

$=\frac {a\sqrt{a+1}-(a+1)\sqrt{a}}{a^2(a+1)-{(a+1)^2}{a}}$

$=\frac {a\sqrt{a+1}-(a+1)\sqrt{a}}{a^3+a^2-a(a^2+2a+1)}$

$=\frac {a\sqrt{a+1}-(a+1)\sqrt{a}}{a^3+a^2-a^3-2a^2-a}$

$=\frac {a\sqrt{a+1}-(a+1)\sqrt{a}}{-a^2-a}$

$=\frac {(a+1)\sqrt{a}-{a}\sqrt{a+1}}{a^2+a}$

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

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

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

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

Therefore, by changing each term of the sum into the final form above, we get a telescoping sum:

$\frac {1}{1\sqrt{2}+2\sqrt{1}} + \frac {1}{2\sqrt{3}+3\sqrt{2}} + \dots + \frac {1}{9999\sqrt{10000}+10000\sqrt{9999}}$

$=\frac {1}{\sqrt{1}} - \frac {1}{\sqrt {2}} + \frac {1}{\sqrt{2}} - \frac {1}{\sqrt {3}} + \dots + \frac {1}{\sqrt{9998}} - \frac {1}{\sqrt {9999}} + \frac {1}{\sqrt{9999}} - \frac {1}{\sqrt {10000}}$

$= \frac {1}{\sqrt {1}} - \frac {1}{\sqrt {10000}}$

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

$= \frac {99}{100}$

Thus, $a+b = 99+100 = \boxed {199}$

The values can be rewritten as functions of $n$ .

$f(n) = \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}$

A summation can be used to find the sum of the expression from $n = 1$ to $n = 9999$ .

$\sum_{n=1}^{9999} f(n)$ $\sum_{n=1}^{9999} \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} = \frac{99}{100}$

So $\frac{99}{100} = \frac{a}{b}$ , where $a=99$ and $b=100$ . $a+b=99+100=\boxed{199}$