Re: Help me with induction
Hello, atomos!
Prove by induction that:
\(\displaystyle \L \sum^n_{r=1} \frac{1}{r(r+1)(r+2)(r+3)}\;=\;\frac{1}{18}\:-\:\frac{1}{3(n+1)(n+2)(n+3)}\)
What is the basic concept of proving by induction?
If you're not familiar with Mathematical Induction, this is a terrible problem to start with!
Verify \(\displaystyle S(1):\L\;\;\frac{1}{1\cdot2\cdot3\cdot4}\:=\:\frac{1}{18}\,-\,\frac{1}{3\cdot2\cdot3\cdot4}\:=\:\frac{1}{24}\) . . . True!
Assume \(\displaystyle S(k)\) is true:
\(\displaystyle \L\;\;\frac{1}{1\cdot2\cdot3\cdot4}\,+\,\frac{1}{2\cdot3\cdot4\cdot5}\,+\,\frac{1}{3\cdot4\cdot5\cdot6}\,+\,\cdots\,+\,\frac{1}{k(k+1)(k+2)(k+3)} \;= \;\frac{1}{18}\,-\,\frac{1}{3(k+1)(k+2)(k+3)}\)
Add \(\displaystyle \,\frac{1}{(k+1)(k+2)(k+3)(k+4)\) to both sides:
\(\displaystyle \L\;\frac{1}{1\cdot2\cdot3\cdot4}\,+\,\frac{1}{2\cdot3\cdot4\cdot5}\,+\,\cdots\,+\,\frac{1}{k(k+1)(k+2)(k+3)}\,+\,\frac{1}{(k+1)(k+2)(k+3)(k+4)}\)
\(\displaystyle \L\;\;\;\;\;= \;\frac{1}{18}\,-\,\frac{1}{3(k+1)k+2)(k+3)} \,+ \,\frac{1}{(k+1)(k+2)(k+3)(k+4)}\)
The left side is the left side of \(\displaystyle S(k+1).\)
We will simplify the right side.
Get a common denominator:
\(\displaystyle \L\;\;\;\;\frac{1}{18}\,-\,\frac{1}{3(k+1)(k+2)(k+3)}\cdot\frac{k+4}{k+4}\,+\,\frac{1}{(k+1)(k+2)(k+3)(k+4)}\cdot\frac{3}{3}\)
\(\displaystyle \L\;\;=\;\frac{1}{18} \,- \,\frac{k\,+\,4}{3(k+1)(k+2)(k+3)(k+4)} \,+ \,\frac{3}{3(k+1)(k+2)(k+3)(k+4)}\)
\(\displaystyle \L\;\;= \;\frac{1}{18} \,+\,\frac{-(k\,+\,4)\,+\,3}{(k+1)(k+2)(k+3)(k+4)} \;= \;\frac{1}{18} \,+ \,\frac{-k\,-\,1}{3(k+1)(k+2)(k+3)(k+4)}\)
\(\displaystyle \L\;\;=\;\frac{1}{18} \,- \,\frac{\sout{k\,+\,1}}{3(\sout{k+1})(k+2)(k+3)(k+4)} \;= \;\frac{1}{18} \,- \,\frac{1}{3(k+2)(k+3)(k+4)}\)
And this is the right side of \(\displaystyle S(k+1).\;\) . . . We're done!
