Product of numbers

$(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{100})$

$=\frac{3}{2}\times\frac{4}{3}...\times\frac{101}{100}$

$=\frac{101}{2}$

$a+b=\boxed{103}$

Note we can rewrite as $(\frac{3}{2})(\frac{4}{3})...(\frac{100}{99})(\frac{101}{100})$ All terms will cancel except the first denominator and the last numerator so it will simplify to $\frac{101}{2}$ Hence solution is $101+2=103$

Firstly, $1+\frac{1}{n}=\frac{n+1}{n}$

Then, we have

$\prod_{n=2}^{N}\frac{n+1}{n}=\frac{3}{2}\times\frac{4}{3}\times...\times\frac{N}{N-1}\times\frac{N+1}{N}=\frac{N+1}{2}$

So, with the given expression, we can rewrite it into as such

$\prod_{n=2}^{100}\frac{n+1}{n}$

which is equivalent to

$\frac{100+1}{2}=\frac{101}{2}$

$\therefore a+b=103$

$(1 + \frac{1}{2})(1 + \frac{1}{3})(1 + \frac{1}{4}).......(1 + \frac{1}{99})(1 + \frac{1}{100})$

$= (\frac{3}{2})(\frac{4}{3})(\frac{5}{4}).....(\frac{100}{99})(\frac{101}{100})$

Everything gets cancelled, only the first denominator term and the last numerator term remain.

Therefore the result $\frac{a}{b} = \frac{101}{2}$

So, $a + b = 101 + 2 = \boxed{103}$

That's the answer!

Firstly we try to observe and identify any pattern that may exist between the terms. On inspection, we find that $t_n$ = $\frac{n+1}{n+2}$ .

Now on evaluating the product for the given sequence, we notice that numerator of the $t_n$ term equals the denominator of the $t_{n-1}$ term; and hence cancel out.Hence the product evaluates to the denominator as the denominator of the first term ie 2, and the numerator as the numerator of the last term, ie 101.

$({1+ \frac{1}{2}}) \times ({1 + \frac{1}{3}}) \times ({1 + \frac{1}{4}}) \times \dots \times ({1 + \frac{1}{100}})$ = $\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{101}{100}$ = $\frac {101}{2}$

(1+1/2)(1+1/3)...(1+1/100) =(3/2)(4/3)(5/4)...(100/99)(101/100). Now you can easily see the the numerator of the first term can be cancelled with the denominator of the next term and goes on. Finally, we have the numerator of the last term(101) and denominator of the first term(2). Hence, 101/2. Therefore, a+b=103.

The general form of each factor is $1 + \frac{1}{n} = \frac {1 + n}{n}$ . So we may infer that the product $( 1 + \frac{1}{2})( 1 + \frac{1}{3})...( 1 + \frac{1}{100}) = (\frac{3}{2})(\frac{4}{3})...(\frac{101}{100}) =$ $= (\frac {2}{2})(\frac{3}{2})(\frac{4}{3})...(\frac{101}{100}) = \frac{101!}{2 \times 100!} =$ $= \frac{101}{2} = \frac {a}{b}$ .

Hence $a + b = 101 + 2 = 103$

Rewriting each term as $\frac{n+1}{n}$ we can see that the numerator of any term is the same as the denominator of the next term, so they cancel each other out. This leaves only the denominator of the first term and the numerator of the last term, which results in the fraction $\frac{101}{2}$ .

$101+2=103$

The answer is 103

3/2x4/3x............101/100=101/2

$= \frac{3}{2}\frac{4}{3}\frac{5}{4} ... \frac{101}{100}$

(1 + 1/2)(1+1/3)(1+1/4).....(1+1/96)(1+1/97)(1 + 1/98)(1 + 1/99)(1 + 1/100) = 2 x 3/2 x 4/3 x ....49/48 x 50/49 x 101/100. There is 49 pair number. So 2 x 3/2 x 4/3 x ....49/48 x 50/49 x 101/100 = 50 x 101/100 = 101 / 2 <=> a = 101, b = 2 <=> a + b = 103.

We operate this multiplication of fraction to be \frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}.....\frac{101}{100} Simplify the denominator of a factor with the nominator of the adjacent factor. And then we get a fraction \frac{101}{2}. And so the answer is 103

Note that $1 + \frac{1}{x+k} = \frac{x+k+1}{x+k}$ , and also that $\frac{x+1}{x} \times \frac{x+2}{x+1} = \frac{x+2}{x}$ .

This product telescopes, and we can note that by multiplying the first terms: $\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \cdots \times\frac{101}{100} = \frac{101}{2}$ . Thus, $a +b = 103.$

(1+1/2)(1+1/3)(1+1/4)...(1+1/100) => (3/2)(4/3)(5/4)... (101/100)

We can clearly see that EVERYTHING CAN BE CANCELLED OUT , Except the DENOMINATOR of the FIRST(3/2) and the NUMERATOR of the LAST(101/100)

So we have; (1/2)(101/1) => 101/2

a+b = 101+2 => 103.

Done :))

Rewrite the expression: Productory of (1+ 1/n) from n=2 to 100. After this, you may realize the numbers above will be divided by the numbers above (in the fractions). Therefore we get: 101/2; where a=101;b=2 ----> a+b=101+2=103

(1+1/2)(1+1/3)... can be written as: 3/2 * 4/3 * 5/4 ...... 101/100. We are simply just writing out each bracket as a fraction. We can now see that most of the integers in the numerator are also in the denominator. We can cancel out all these numbers. In fact all numbers cancel out apart from 101 in the numerator and 2 in the denominator. Thus you get 101/2. And so 101+2=103

this can be rewritten as 3/2 * 4/3 * 5/4 * ... * 101/100

All the denominators and dividers cancel out, since a/b*b/c=a/c if b and c are real and unequal to zero, except for 101 and 2, so a/b = 101/2

101 and 2 are coprime positive integers and cannot be further simplified, so a=101 and b=2, hence a+b=103