Deciphering an equation

Algebra Level 4

Assume we have series $f(1), f(2), \ldots , f(n)$ , where $f(i)$ is a function defined for natural numbers. We know that this is true

$\sum _{ k=1 }^{ n }{ f(k) } =f(n)(n+1)-n$

Find the function $f(n)$

f(n)={ 2 }^{ n+2 }

f(n)=\sum _{ i=1 }^{ n }{ \frac { i }{ \left( 2i+5 \right) ! } }

f(n)=1-\sum _{ i=1 }^{ n }{ \frac { 1 }{ i\left( i-1 \right) } }

f(n)=\sum _{ i=1 }^{ n }{ \frac { 1 }{ i } }

1 solution

Karen Vardanyan
Jun 6, 2014

We have this equation:

$\sum _{ k=1 }^{ n }{ f(k) } =f(n)(n+1)-n$ .

Now let's write this equation for $n-1$ :

$\sum _{ k=1 }^{ n-1 }{ f(k) } =f(n-1)n-(n-1)$ .

Now, if we subtract the second equation from the first one, after the simplifications we'll get this:

$f(n)-f(n-1)=\frac { 1 }{ n }$ .

Thus we get for $f(n)$ :

$f(n)=\sum _{ i=1 }^{ n }{ \frac { 1 }{ i } }+C$ .

It's quite easy to check that only in the case of $C=0$ our equation will be true. So we find the wanted function:

$f(n)=\sum _{ i=1 }^{ n }{ \frac { 1 }{ i } }$