Function and Triangle are Best Friend

$\begin{aligned} f(1)+f(2)+f(3)+…+f(n-1)+f(n)&=n^2 f(n) \\ f(1)+f(2)+f(3)+…+f(n-1) &= (n-1)^2 f(n-1) \\ \implies f(n)&= n^2 f(n)-(n-1)^2 f(n-1) \\ n^2 f(n) - f(n)&= (n-1)^2 f(n-1) \\ f(n) (n^2-1)&=(n-1)^2 f(n-1)\\ f(n) &= \frac{(n-1)^2 f(n-1)}{(n^2 -1)} \\ f(n) &= \frac{(n-1)(n-1)f(n-1)}{(n+1)(n-1)} \\ f(n) &= \frac{(n-1)f(n-1)}{(n+1)}\\ \end{aligned}$

So $f(n) = \frac{n-1}{n+1} \times f(n-1)$ $\begin{aligned} f(1) &= \frac{2017}{1}\\ f(2) &= \frac{2017}{3}\\ f(3) &= \frac{2017}{6}\\ f(4)& = \frac{2017}{10}\\ f(5) &= \frac{2017}{15}\\ f(6) &= \frac{2017}{21}\\ \end{aligned}$ Notice how the denominator was in fact triangular numbers, with $a_n = \frac{n(n+1)}{2}$ . So,

$f(2017) = \large \frac{2017}{\frac{2017(2018)}{2}} = 2017 \times \frac{2}{2017(2018)} = \frac{2}{2018} = \color{#D61F06}{\boxed{\frac{1}{1009}}}$

Let $\displaystyle S_n = \sum_{k=1}^n f(k)$ , then we have:

$\begin{aligned} S_n & = n^2 f(n) \\ S_{n-1} + f(n) & = n^2 f(n) \\ S_{n-1} & = (n^2-1) f(n) \\ \implies S_n & = ((n+1)^2 -1) f(n+1) \\ S_{n-1} + f(n) & = (n^2+2n)f(n+1) \\ (n^2-1) f(n) + f(n) & = n(n+2)f(n+1) \\ \implies f(n+1) & = \frac {nf(n)}{n+2} \end{aligned}$

The first few $n$ shows that $f(n) = \dfrac {4034}{n(n+1)}$ . Let prove by induction that the claim is true for all $n \ge 1$ .

The Proof

• For $n=1$ , $f(1) = \dfrac {4034}{1\cdot 2} = 2017$ as given, therefore, the claim is true for $n=1$ .

• Assuming the claim is true for $n$ , then:

$\begin{aligned} \quad f({\color{#3D99F6}n+1}) & = \frac {n(f(x)}{n+2} \\ & = \frac n{n+2} \cdot \frac {4034}{n(n+1)} \\ & = \frac {4034}{({\color{#3D99F6}n+1)}({\color{#3D99F6}n+1}+1)} \end{aligned}$

$\quad$ The claim is also true for $n+1$ , therefore, the claim is true for all $n \ge 1$ .

Now, we have $f(2017) = \dfrac {4034}{2017\cdot 2018} = \dfrac 1{1009}$ . Therefore, $a+b = 1 + 1009 = \boxed{1010}$ .

$\\ f(1)+f(2)+f(3)+...+f(n)={ n }^{ 2 }f(n)\\ { \left( n-1 \right) }^{ 2 }f(n-1)+f(n)={ n }^{ 2 }f(n)\\ \\ { \left( n-1 \right) }^{ 2 }f(n-1)=\left( n-1 \right) \left( n+1 \right) f(n)\\ (n-1)f(n-1)=(n+1)f(n)\\ \frac { f(n) }{ f(n-1) } =\frac { n-1 }{ n+1 } \\ \\ \frac { f(2) }{ f(1) } \cdot \frac { f(3) }{ f(2) } \cdot \frac { f(4) }{ f(3) } \cdot \cdots \cdot \frac { f(2016) }{ f(2015) } \cdot \frac { f(2017) }{ f(2016) } =\frac { f(2017) }{ f(1) } =\frac { f(2017) }{ 2017 } \\ \\ f(2017)=2017\cdot \frac { 1 }{ 3 } \cdot \frac { 2 }{ 4 } \cdot \frac { 3 }{ 5 } \cdot \cdots \cdot \frac { 2015 }{ 2017 } \cdot \frac { 2016 }{ 2018 } =\frac { 2017\cdot 1\cdot 2 }{ 2017\cdot 2018 } =\frac { 1 }{ 1009 } \\ \\ \boxed { \quad \therefore \quad a+b=1+1009=1010\quad } \\$