Linear cubics

The general term of this expression is $\dfrac{2n-1}{(n-1)n(n+1)}.$ We simplify this expression so as to get a telescoping series: $\begin{aligned} \dfrac{2n-1}{(n-1)n(n+1)} &=& \frac{n}{(n-1)n(n+1)} + \frac{n-1}{(n-1)n(n+1)} \\ &=& \frac{1}{(n-1)(n+1)} + \frac{1}{n(n+1)} \\ &=& \dfrac{\frac{1}{n-1} - \frac{1}{n+1}}{2} + \frac{1}{n} - \frac{1}{n+1}. \end{aligned}$ Now, we substitute in the specific terms into our general expression to get $\dfrac{\frac11 - \frac13 + \frac12 - \frac14 + \cdots + \frac1{40} - \frac1{42}}{2} + \frac12 - \frac13 + \frac13 - \frac14 + \cdots + \frac1{41} - \frac1{42}.$ After canceling like terms, we are left with $\begin{aligned} \dfrac{1 + \frac12 - \frac1{41} - \frac1{42}}2 + \frac12 - \frac1{42} &=& \frac12 + \frac12 + \frac14 - \frac1{42} - \frac1{82} - \frac1{84} \\ &=& 1 - \frac1{82} + \frac{18}{84} \\ &=& \frac{4140}{3444} \\ &=& \frac{345}{287}. \end{aligned}$ Hence, our answer is $345 + 287 = \boxed{632}.$

The expression is equivalent to $\sum_{n=2}^{n=41}\frac{2n-1}{(n-1)(n)(n+1)}$ and by partial fraction decomposition we see that it is also equivalent to $\sum_{n=2}^{n=41}\frac{\frac{1}{2}}{n-1}+\frac{1}{n}+\frac{-\frac{3}{2}}{n+1}$ Expanding the first few terms we notice for three consecutive $n$ ; $k$ , $k+1$ , and $k+2$ ; we have $\frac{\frac{1}{2}}{(k+2)-1}+\frac{1}{k+1}+\frac{-\frac{3}{2}}{k+1}=0$ Thus, the series telescopes. After all cancellations are made, the remaining terms are $\frac{\frac{1}{2}}{2-1}+\frac{1}{2}+\frac{\frac{1}{2}}{3-1}+\frac{-\frac{3}{2}}{40+1}+\frac{1}{41}+\frac{-\frac{3}{2}}{41+1}$ Which is equivalent to $\frac{1}{2}+\frac{3}{4}-\frac{1}{82}-\frac{3}{84}=\frac{345}{287}$ Thus, the answer is $345+287=632$

$\displaystyle \sum_{r = 1}^{40} \frac{2r + 1}{r(r + 1)(r + 2)}$

= $\displaystyle \sum_{r = 1}^{40} \frac{(r + 1)(r + 2) - r(r + 1) - 1}{r(r + 1)(r + 2)}$

= $\displaystyle \sum_{r = 1}^ {40} \bigg ( \frac{1}{r} - \frac{1}{r + 2}\bigg) - \sum_{r = 1}^{40} \frac{1}{r(r + 1)(r + 2)}$

= $\displaystyle \sum_{r = 1}^{40} \bigg(\frac{1}{r} - \frac{1}{r + 1}\bigg) +\sum_{r=1}^{40}\bigg(\frac{1}{r + 1} - \frac{1}{r + 2} \bigg)$ - $\frac{1}{2} \displaystyle \sum_{r = 1}^{40}\Bigg( \frac{1}{r(r + 1)} - \frac{1}{(r +1)(r + 2)}\Bigg)$

= $1 - \frac{1}{41} + \frac{1}{2} - \frac{1}{42} - \frac{1}{4} + \frac{1}{2} \frac{1}{41}{42} = \frac{345}{287}$

$\Rightarrow a + b = \fbox{632}$

We decompose into partial fractions that then telescope, so $\begin{aligned} \sum_{i = 1}^{40} \frac{2i + 1}{i(i + 1)(i + 2)} &= \sum_{i = 1}^{40} \Bigg(\frac{1}{i + 1} + \frac{1}{2i} - \frac{3}{2(i + 2)}\Bigg) \\ &= \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{41} - \frac{3}{82} - \frac{3}{84} \\ &= \frac{345}{287} \end{aligned}$ and hence $a + b = 345 + 287 = \boxed{632}$

The partial fractions decomposition of the general term $\frac{2n+1}{n(n+1)(n+2)}$ has the form $\frac{2n+1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}$ . Multiplying both sides by $n(n+1)(n+2)$ gives us $2n+1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1)$ . Substituting $n = 0$ gives us $A = \frac{1}{2}$ ; substituting $n = -1$ gives us $B = 1$ ; and substituting $n = -2$ gives us $C = -\frac{3}{2}$ . Therefore $\frac{2n+1}{n(n+1)(n+2)} = \frac{1}{2n} + \frac{1}{n+1} - \frac{3}{2(n+2)}$ and our sum can be written as $\left( \frac{1}{2} + \frac{1}{2} - \frac{3}{6} \right) + \left( \frac{1}{4} + \frac{1}{3} - \frac{3}{8} \right) + \left( \frac{1}{6} + \frac{1}{4} - \frac{3}{10} \right) + \cdots + \left( \frac{1}{2(40)} + \frac{1}{41} - \frac{3}{2(42)} \right).$ Note that $\frac{3}{2(n+2)} = \frac{1}{n+2} + \frac{1}{2(n+2)}$ , so that the last term in each triple $-\frac{3}{2(n+2)}$ cancels with the middle term in the next triple $\frac{1}{n+2}$ and the first term of the triple after that $\frac{1}{2(n+2)}$ . The terms that do not cancel are the first two terms in the first triple, the first term in the second triple, the last term in the next-to-last triple, and the last two terms in the last triple. Hence, our sum equals $\left( \frac{1}{2} + \frac{1}{2} \right) + \left( \frac{1}{4} \right) + \left( -\frac{3}{2(41)} \right) + \left( \frac{1}{41} - \frac{3}{2(42)} \right).$ In reduced form this is equal to the fraction $\frac{345}{287}$ , and thus $a + b = 345 + 287 = 632$ .

It can be written as:

= ( 1/2-1/3+1/3-1/4.......1/41-1/42 ) + 1/2[1/2-1/(41x42)]

= 20/21 + 215/861

= 1035/861 = 345/287

Ans. = 345 + 287 = 632

Note que essa soma pode ser escrita como o seguinte somatório: £(2n+1/n(n+1)(n+2)) de n igual 1 até 40

Esse somatório pode ser separado em dois:
£(1/(n+1)(n+2)) + £(1/n(n+2))
£(1/(n+1) - 1/(n+2)) + 1/2 £(1/n - 1/(n+2))

Analisando o primeiro somatório temos:
(1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) +...+ (1/41 - 1/42) = 1/2 - 1/42 = 10/21

Analisando o segundo somatório:
1/2 [(1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) +...+ (1/40 - 1/42)] =
= 1/2 (1 + 1/2 - 1/41 - 1/42) = 625/41×3×7

Assim:
£(2n + 1/n(n+1)(n+2)) = 5×2/3×7 + 5^4/3×7×41 = 345/287

By partial fraction the nth term is
(2n - 1)/{(n-1)n(n+1)= 1/{2(n-1)} + 1/n - (3/2)/(n+1)
= Sum{ ( 1/2 )/(n-1) -(1/2)/n } + Sum{ 1/n - 1/(n+1) } , n = 1 to 41
Telescopic series,
= {1/2 + 1/4 - 1/82 - 1/84 } + {1/2 -1/42 }
=345/287 =a/b
a + b =632

We can simplify this problem by using Partial Factor Decomposition

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

\frac{5}{\( 2 \times \( 3 \times 4 )} ) = $\frac{1}{4}$ + $\frac{1}{3}$ - $\frac{3}{8}$ ...

By continuing this process we can arrive to the result

$\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{6}$ + ... + $\frac{1}{80}$ + $\frac{2}{4}$ + $\frac{2}{6}$ + $\frac{2}{8}$ + ... + $\frac{2}{41}$ - $\frac{3}{6}$ - $\frac{3}{8}$ - $\frac{3}{10}$ - ... - $\frac{3}{84}$

which can be simplified into

$\frac{1}{2}$ (1 + \frac{1}{2} \ + \( \frac{1}{3} \ + ... + \( \frac{1}{40} \+ 2( \( \frac{1}{2} \ + \( \frac{1}{3} \ + \( \frac{1}{4} \ \( \frac{1}{5} \ + ... + \( \frac{1}{41} \ ) - 3( \( \frac{1}{3} \ + \( \frac{1}{4} \ + \( \frac{1}{5} \ + ... + \( \frac{1}{42} ))

which is also equal to

$\frac{1}{2}$ ( 1 + $\frac{3}{2}$ - $\frac{1}{41}$ - $\frac{3}{41}$ )

which is equal to $\frac{345}{287}$

therefore a = 345 and b = 287, a + b = 632