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For $n \ge 1$, let
$$\frac{1}{1 \cdot 5} + \frac{1}{5 \cdot 9} + \cdots + \frac{1}{(4n - 3)(4n + 1)}$$
Guess a simple explicit formula for $a_n$ and prove it by induction.

How do I actually guess a explicit formula for $a_n$? I'm confused, can anyone show me how to solve?

jhg
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  • See https://math.stackexchange.com/questions/2284350/guess-the-formula-and-prove-by-induction – lab bhattacharjee May 18 '17 at 05:48
  • Use partial fractions! – Saketh Malyala May 18 '17 at 05:48
  • See also https://math.stackexchange.com/questions/2284505/induction-provide-formula-prove-it-by-induction/2284514#2284514 – Robert Z May 18 '17 at 05:52
  • Guessing usually consists of trying values of $n$ that are particularly nice. For example, $n=1$ gives $\frac 1 {5}$, $n=2$ gives $\frac 2 9$, $n=3$ gives $\frac{3}{13}$. See if you can find a pattern to test on $n=5$ or more. – Artimis Fowl May 18 '17 at 05:54
  • I edited the title of the post with answers, to make it more visible. – zwim May 18 '17 at 05:55
  • Guessing is... guessing. You try a few and see what happens. It's inefficient but a good thing is you might see a patter. And a trick to solving. a_1= 1/5 answer a_2 =1/5 +1/5.9=9/5.9 + 1/5.9 =10/45=2/9. a_3 =2/9+1/9.13=27/9.13=3/13. Should be a strong pattern emmerging. – fleablood May 18 '17 at 06:09

2 Answers2

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Let the given statement be $P(n):\Large {1 \over 1.5} + {1 \over 5.9}+....+{1 \over (4n-3).(4n+1)}$

$$\sum_{r=1}^{n} {1 \over (4r-3)(4r+1)} $$

$$\frac{1}{4} \sum_{r=1}^{n} {1 \over (4r-3)} -{1 \over (4r+1)} $$

$$\frac{1}{4} \left(1-\frac{1}{4n+1} \right)$$

$P(1)=\frac{1}{1.5}$. Hence $P(1)$ is true.

Assuming $P(k)$ is true where $k \ge 1$

We will try to prove $P(k+1)$ is true

$$P(k+1)= {1 \over 1.5} + {1 \over 5.9}+....+{1 \over (4k-3).(4k+1)} + {1 \over (4(k+1)-3).(4(k+1)+1)}$$

$$P(k+1)= {1 \over 1.5} + {1 \over 5.9}+....+{1 \over (4k-3).(4k+1)} + {1 \over (4k+1).(4k+5)}$$

$$\frac{1}{4} \left(1-\frac{1}{4k+1} \right)+{1 \over (4k+1).(4k+5)}$$

$$\frac{1}{4} -\frac{1}{4k+1}\left(\frac{1}{4}-\frac{1}{4k+5} \right)$$ $$\frac{1}{4} -\frac{1}{4k+1}\left(\frac{4k+1}{4(4k+5)} \right)$$

$$\frac{1}{4} -\left(\frac{1}{4(4k+5)} \right)$$

$$\frac{1}{4}\left(1 -\frac{1}{(4k+5)} \right)$$

Thus $P(k+1)$ is true whenever $P(k)$ is true.

Aman Sharma
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Yu already have an explicit formula which is $$\begin{array}{lcl} a_1 & = & \frac{2}{9} \\ \vdots\\ a_n & = & \frac{1}{(4n-3)(4n+1)}=\frac{1}{16n^2-8n-3} \\ \end{array}$$

It suffices to show that this is true, using induction.

Parseval
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