The next year summation

Try a slightly harder version here ..

$\large\mathfrak{S}=\displaystyle\sum_{n=1}^{2017}\left(\dfrac{n}{\displaystyle\sum_{r=1}^n r(r+1)}\right)$ Using formula for sum of n,n $^2$ , denominator can be simplified as: $\large\displaystyle\sum_{n=1}^{2017}\left(\dfrac{\not n}{\frac{\not n(n+1)(2n+1)}{6}+\frac{\not n(n+1)}{2}}\right)$ $\large=\displaystyle\sum_{n=1}^{2017}\left(\dfrac{3}{(n+1)(n+2)}\right)$ Using Partial Fractions-Cover Up rule :- $\dfrac{1}{(n+1)(n+2)}=\dfrac{1}{n+1}-\dfrac{1}{n+2}$ Therefore, $\large\mathfrak S=3\displaystyle\sum_{n=1}^{2017}\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)$ $\large\color{#69047E}{\mathbf{A Telescopic Series}}$ $\large =3\left(\dfrac{1}{2}-\dfrac{1}{2019}\right)$ $\large =\dfrac{2017}{1346}$ $\Huge\therefore 2017+1346=\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{3363}}}}}$

$\begin{aligned} & \displaystyle \sum_{n=1}^{2017} \dfrac{n}{1\cdot2+2\cdot3+\ldots+n(n+1)} \\ =& \sum_{n=1}^{2017} \dfrac{n}{\frac{n(n+1)(n+2)}{3}} \quad \quad [\small{\text{See Note}}] \\ =&3\sum_{n=1}^{2017} \dfrac{1}{(n+1)(n+2)} \\ =& \sum_{n=1}^{2017} \dfrac{1}{n+1}-\dfrac{1}{n+2} \quad \quad [\text{Telescopic Series}] \\ =& 3\left[\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\ldots-\dfrac{1}{2019}\right] \\ =& 3\left[\dfrac{1}{2}-\dfrac{1}{2019}\right] \\ =& \boxed{\dfrac{2017}{1346}} \end{aligned}$

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Note : We have $1\cdot2+2\cdot3+\ldots+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$

Here's a proof by Induction .

Proof : Since $1\cdot2=\dfrac{1\cdot2\cdot3}{3}=2$ ,the statement holds for $n=1$ .

Now let's assume $1\cdot2+2\cdot3+\ldots+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$ .Then add $(n+1)(n+2)$ to both sides to get $\begin{aligned} 1\cdot2+2\cdot3+\ldots+n(n+1)+(n+1)(n+2) =& \dfrac{n(n+1)(n+2)}{3} +(n+1)(n+2) \\ =&(n+1)(n+2)\left[\dfrac{n}{3}+1\right] \\ =&\dfrac{(n+1)(n+2)(n+3)}{3} \end{aligned}$

Thus if it holds for $n$ ,it also holds for $n+1$ .But it is true for $1$ ,thus it is true for $2$ ,thus it is true for $3$ ,and so on.Therefore the statement holds universally.

Ways to finish the summation have been provided elsewhere, but here's a better way for the denominator:

$\displaystyle\begin{aligned} 1 \cdot 2 + 2 \cdot 3 + \ldots + n \cdot (n+1) &= 2 \left( \frac{1 \cdot 2}{2} + \frac{2 \cdot 3}{2} + \ldots + \frac{n \cdot (n+1)}{2} \right) \\ &= 2 \left( \binom{2}{2} + \binom{3}{2} + \ldots + \binom{n+1}{2} \right) \\ &= 2 \binom{n+2}{3} \\ &= 2 \cdot \frac{n(n+1)(n+2)}{6} \\ &= \frac{n(n+1)(n+2)}{3} \end{aligned}$

The second to third line is because of the identity $\binom{k}{k} + \binom{k+1}{k} + \binom{k+2}{k} + \ldots + \binom{n}{k} = \binom{n+1}{k+1}$ . This can be proven by noticing $\binom{k}{k} = 1 = \binom{k+1}{k+1}$ and then using the identity $\binom{a}{b} + \binom{a}{b+1} = \binom{a+1}{b+1}$ repeatedly.