Sum of Fractions

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

$\Rightarrow a_{n}=\frac {1}{n+2}+\frac {1}{n(n+2)}-\frac {1}{n+1}=\frac {1}{n+2}-\frac {1}{n+1}+\frac {1}{2}(\frac {1}{n}-\frac {1}{n+2})$

$S=\displaystyle \sum_{n=1}^{14} a_{n}= \displaystyle \sum_{n=1}^{14}(\frac {1}{n+2}-\frac {1}{n+1})+\frac {1}{2}\displaystyle \sum_{n=1}^{14}(\frac {1}{n}-\frac {1}{n+2})$

$\Rightarrow S=(\frac {1}{16}-\frac {1}{2})+\frac {1}{2}(1+\frac {1}{2}-\frac {1}{15}-\frac {1}{16})$

$\Rightarrow S=\frac {119}{480} \Rightarrow a=119;b=480 \Rightarrow \boxed{a+b=599}$

We will prove a formula over all $\displaystyle{\sum_{i=1}^k{\frac{1}{i(i+1)(i+2)}}}$ , generalizing the problem.

We hope that the sum of the sequence telescopes, or that the terms will conveniently collapse so that we can find the sum easily. To do this, we first try to write the value $\frac{1}{i(i+1)(i+2)}$ as $\frac{a}{i}+\frac{b}{i+1}+\frac{c}{i+2}$ , a partial fraction decomposition. The motivation for doing this lies in the fact that expressing each term as a sum of fractions may lead to our supposed telescoping.

So $\frac{1}{i(i+1)(i+2)}=\frac{a}{i}+\frac{b}{i+1}+\frac{c}{i+2}$ over all $i$ . We can plug in any values for $i$ to solve for $a,b$ and $c$ . Plugging in $i=0$ , we get that $a=1/2$ . Plugging in $i=-1$ , we get that $b=-1$ . Finally, plugging in $i=-2$ , we get $c=1/2$ .

So, each value of $\frac{1}{i(i+1)(i+2)}$ can be written as $\frac{1}{2} \cdot \frac{1}{k}-\frac{1}{k+1}+\frac{1}{2} \cdot \frac{1}{k+2}$ . We now express the sum with this decomposition: $\displaystyle{\sum_{i=1}^k{\frac{1}{i(i+1)(i+2)}}}$ $=(\frac{1}{2} \cdot \frac{1}{1}-\frac{1}{2}+\frac{1}{2} \cdot \frac{1}{3})+(\frac{1}{2} \cdot \frac{1}{2}-\frac{1}{3}+\frac{1}{2} \cdot \frac{1}{4})+(\frac{1}{2} \cdot \frac{1}{3} -\frac{1}{4}+\frac{1}{2} \cdot \frac{1}{5})+$ $(\frac{1}{2} \cdot \frac{1}{4}-\frac{1}{5}+\frac{1}{2} \cdot \frac{1}{6}) \cdots$ and so on.

Notice that we can pair two $\frac{1}{2} \cdot \frac{1}{3}$ to make a total $\frac{1}{3}$ , which cancels with the $-\frac{1}{3}$ , and so on for $3, 4, 5 \cdots$ .

So our final value is simply $\frac{1}{2} \cdot \frac{1}{1}-\frac{1}{2} \cdot \frac{1}{2}-\frac{1}{2} \cdot \frac{1}{i+1}+\frac{1}{2} \cdot \frac{1}{i+2}$ . Simplifying leads to $\frac{1}{4}-\frac{1}{2(i+1)}+\frac{1}{2(i+2)}$ , or more simply just $\frac{n(n+3)}{4(n+2)(n+1)}$ .

We have our general formula. Plugging in $i=14$ leads to $\frac{119}{480}$ , so the desired answer is $599$ .

Let $u_i=\dfrac{1}{i(i+1)},a_n=\dfrac{1}{n(n+1)(n+2)}$ Then, $u_i-u_{i+1}=2a_i$ So, telescoping, the sum is just \dfrac12\left(u_1-u_{14}\right)=\dfrac12\left(\dfrac{1}{2}-\dfrac{1}{15*16}\right=\dfrac{119}{480}) Since we already have $\gcd(480,119)=1$ , the sum is $119+480=\boxed{599}$

We write

$\frac{1}{(n)(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}$

Where $A,B,C\in \ \mathbb{R}$ . Combining the fractions on the RHS gives

$\frac{1}{(n)(n+1)(n+2)} = \frac{A(n+1)(n+2) +B(n)(n+2) +C(n)(n+1)}{(n)(n+1)(n+2)}$

$\Rightarrow A(n^2+3n+2) + B(n^2+2n) + C(n^2+n) = 1$

The RHS of the last equation only has a constant. We thus obtain the following systems of equations

$An^2 + Bn^2 +Cn^2 = 0 \Rightarrow A+B+C = 0$

$3An + 2Bn + Cn = 0 \Rightarrow 3A+2B+C=0$

$2A = 1 \Rightarrow A=\frac{1}{2}$

Solve for $B$ and $C$ gives $(A,B,C) = (\frac{1}{2}, -1, \frac{1}{2} )$ . The summation can then be written as

$\displaystyle \sum^{14}_{i=1} \left(\frac{\frac{1}{2}}{i} - \frac{1}{i+1} + \frac{\frac{1}{2}}{i+2}\right)$

$\displaystyle = \frac{\frac{1}{2}}{1} - \frac{1}{2} +\frac{\frac{1}{2}}{3}$

$\displaystyle+\frac{\frac{1}{2}}{2} - \frac{1}{3} +\frac{\frac{1}{2}}{4}$

$\displaystyle+ \frac{\frac{1}{2}}{3} - \frac{1}{4} + \frac{\frac{1}{2}}{5}$

$\displaystyle + \ \ \vdots + \ \ \ \vdots + \ \vdots$

$\displaystyle + \frac{\frac{1}{2}}{13} - \frac{1}{14} + \frac{\frac{1}{2}}{15}$

$\displaystyle +\frac{\frac{1}{2}}{14} - \frac{1}{15} + \frac{\frac{1}{2}}{16}.$

This summation telescopes, leaving $S=\frac{\frac{1}{2}}{1} - \frac{1}{2} + \frac{\frac{1}{2}}{2} + \frac{\frac{1}{2}}{15} -\frac{1}{15} + \frac{\frac{1}{2}}{16} = \frac{119}{480}$ . Our answer is $119+480 = 599$ .

Solution for Math Question Observe that 1/(n×(n+1)×(n+2))=1/n (1/(n+1)-1/(n+2))=1/(n+2) (1/n-1/(n+1)) As such, we have S=1/1 (1/2-1/3)+1/2 (1/3-1/4)+⋯+1/14 (1/15-1/16) =(1/1×1/2)-(1/1×1/3)+(1/2×1/3)-(1/2×1/4)+⋯+(1/14×1/15)-(1/14×1/16) =(1/1×1/2)+1/3 (1/2-1/1)+1/4 (1/3-1/2)+⋯+1/15 (1/14-1/13)-(1/14×1/16) =(1/1×1/2)-1/3 (1/1-1/2)-1/4 (1/2-1/3)-…-1/15 (1/13-1/14)-(1/14×1/16) =(1/1×1/2)-1/(1×2×3)-1/(2×3×4)-…-1/(13×14×15)-(1/14×1/16) =1/2-1/(1×2×3)-1/(2×3×4)-…-1/(13×14×15)-1/(14×15×16)+1/(14×15×16)-1/(14×16) =1/2-S+1/(14×15×16)-1/(14×16) Hence, 2S=1/2+1/(14×15×16)-1/(14×16) =1/2+(1-15)/(14×15×16) =1/2-1/(15×16) =1/2-1/15+1/16 =119/240 Resulting in, S=119/480

we can observe that nth term is 1/n(n+1)(n+2). Let V(n) be 1/n(n+1). Then if we do v(n) - v(n+1), we get 2/n(n+1)(n+2). which is equal to 2.T(n) therefore t(n) = [v(n) - v(n+1)]/2. when we do t(1) + t(2) + ........... + t(14) we get [v(1) - v(15)]/2 v(1) = 1/2 and v(15) = 1/240. and then on further simplification we get the answer as 119/480. therefore a=119 and b=480 and a + b = 119+480 = 599

Trying to calculate this sum using brute force is far too tedious, thus we must find a way to simplify this problem.

Making use of the fact that $\frac{1}{k}-\frac{1}{k+2}=\frac{1}{2} \frac{1}{k(k+2)}$ , we can rewrite this sum as:

$\frac{1}{2}\frac{1}{n+1}(\frac{1}{n}-\frac{1}{n+2})+\frac{1}{2}(\frac{1}{n+2}\frac{1}{n+1}-\frac{1}{n+3})+ \ldots +\frac{1}{2}\frac{1}{n+14}(\frac{1}{n+13}-\frac{1}{n+15})$ $= \frac{1}{2n(n+1)}-\frac{1}{2(n+1)(n+2)}+\frac{1}{2(n+1)(n+2)}-\frac{1}{2(n+2)(n+3)}+ \ldots +\frac{1}{2(n+13)(n+14)}-\frac{1}{2(n+14)(n+5)}$ $=\frac{1}{2n(n+1)}-\frac{1}{2(n+14)(1+15)}$

Substituting n=1, we get $S=\frac{1}{4}-\frac{1}{480}=\frac{119}{480}$ Hence, $a+b=599$

Firstly, we realise that S is the difference between the sequence

M = \frac{1}{1\times2} + \frac{1}{2\times3} + ... + \frac{1}{14\times15}

and

N = \frac{1}{1\times3} + \frac{1}{2\times4} + ... + \frac{1}{14\times16}

Then we calculate the value of M and N

M = \frac{1}{1\times2} + \frac{1}{2\times3} + ... + \frac{1}{14\times15} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ... + (\frac{1}{14} - \frac{1}{15}) = 1 - \frac{1}{15} =\frac{14}{15}

N = \frac{1}{1\times3} + \frac{1}{2\times4} + ... + \frac{1}{14\times16} = \frac{1}{2} ( (\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + ... + (\frac{1}{14} - \frac{1}{16}) ) =\frac{1}{2} ( 1 +\frac{1}{2} - \frac{1}{15} -\frac{1}{16}) = \frac{329}{480}

M - N = \frac{14}{15} - \frac{329}{480} = \frac{119}{480}

Hence, a = 119, b = 480 and a + b = 599 (solved!)

1/(n) (n+1) (n+2) = 1/2 * (1/n-1/n+1-1/n+1+1/n+2)

So that, = 1/2 * (1/1-1/2-1/2+1/3)+1/2(1/2-1/3-1/3+1/4)+.....+1/2(1/14-1/15-1/15+1/16)

= 1/2 (1/2-1/15 16) = 1/2 ((120-1)/240) =1/2 (119/240) =119/480

Since a = 119 and b = 480, a + b = 599

1/(n)(n+1)(n+2) = (1/2)(2/(n)(n+1)(n+2)) = (1/2)((n+2-n)/(n)(n+1)(n+2)) This equals (1/2)(1/(n)(n+1) - 1/(n+1)(n+2) ).

So 1/(1)(2)(3) = (1/2)(1/(1)(2)-1/(2)(3)) 1/(2)(3)(4) = (1/2)(1/(2)(3)-1/(3)(4)) ...... 1/(14)(15)(16) = (1/2)(1/(14)(15)-1/(15)(16))

This sum is = (1/2)(1/(1)(2)-1/(2)(3) +1/(2)(3)-1/(3)(4)+...+1/(14)(15)-1/(15)(16))

Cancelling out similar terms, like -1/(2)(3) +1/(2)(3) = 0, we get (1/2)(1/(1)(2)-1/(15)(16)) = (1/2)(1/2-1/240) = 119/480

119 = 7 17, 480 = 2^5 *3 5 in prime factorisation. 119 and 480 are positive, coprime integers. 119 + 480 = 599 (Ans.)

Tn=0.5*(1/n(n+1)-1/(n+1)(n+2)) Now put value for n in this. And start adding u will see that last part of first one get cancelled by first term of succeding one and finally this will reduce onluy to two part.

Note that 1/[(i)(i+1)(i+2]=1/(2i)-1/(i+1)+1/[2(i+2)]. 2/[(i)(i+1)(i+2)]=[1/i-1/(i+1)]-[1/(i+1)-1/(i+2)]. 2S=[(1/1-1/2)-(1/2-1/3)]+[(1/2-1/3)-(1/3-1/4)]+...+[(1/14-1/15)-(1/15-1/16)]=(1/1-1/2+1/2-1/3+...+1/14-1/15)-(1/2-1/3+1/3-1/4+...+1/15-1/16)=(1/1-1/15)-(1/2-1/16)=14/15-7/16=199/240. Therefore, S=199/480. a+b=199+480=599.

This sum is complicated to do directly, but we may rearrange some terms of each fraction separately. As there are three terms in each denominator, the telescoping series may be constructed by using partial sums that have only two terms in their denominator. Therefore, we are looking for a expression like this:

$\frac {1}{(n)(n+1)(n+2)} = A \left( \frac {1}{(n)(n+1)} - \frac {1}{(n+1)(n+2)} \right),$

where A is a constant number. Solving for every possible n , we may find that:

$A = \frac {1}{2} .$

Returning to the first sum S , using the property above:

$S = \frac{1}{2} \left( \frac{1}{1 \times 2} - \frac{1}{2 \times 3} + \frac{1}{2 \times 3} - \frac{1}{3 \times 4} + ... + \frac{1}{14 \times 15} - \frac{1}{15 \times 16} \right).$

Regrouping the terms:

$S= \frac{1}{2} \left( \frac{1}{1 \times 2} - \frac{1}{15 \times 16} \right) = \frac{119}{480} = \frac{a}{b}.$

As: $119 = 7 \times 17$ and $480 = 32 \times 3 \times 5$ , we have that 119 and 480 are coprime integers. Therefore, $a = 119, b = 480$ and: $a+b = 599$ , the solution of this problem.

$S = \frac {1}{1\times 2 \times 3} + \ldots + \frac {1} {(n)(n+1)(n+2)} + \ldots +\frac {1}{14 \times (14+1) \times (14+2)} =$ $= \displaystyle \sum_{n=3}^{16} \frac {(n-3)!}{n!} =$ $\frac {16^2 - 16 -2}{4 \times 16(16-1)} = \frac {119}{480} = \frac {a} {b}$

$a+b = 599$

this is of the form summation(1/n+1/(n+1)+1/(n+2)). so, by partial fraction method and expanding it comes finally to this form- 1/2summation(1/n - 1/n+1) -1/2summation(1/n+1 - 1/n+2). which is solved by taking n=1,2,3....14... we find that only the 1st and last terms in each bracket remains.

taking the given equation into the form telescopic series we get the answer as 199/480.therefore (a+b)=599