Orta ogretim icin zor bir soru, katagori Lisans olmali.
Gozlem: $\dfrac{5}{4}+\dfrac{13}{12}+\dots,\dfrac{841}{840}=1+\dfrac{1}{4}+1+\dfrac{1}{12}+\dots,1+\dfrac{1}{840}$
$1+1+1+\dots+1+\dfrac{1}{4}+\dfrac{1}{12}+\dots,\dfrac{1}{840}$
Dizimiz bu olsun $4,12,24,40,\dots,840$. Farklarini alalim.
$8,12,16,\dots$. Bir daha farklarini alalim.
$4,4,\dots$ ikinci farklar sabit cikti.
Demek ki dizimiz $s(n)=an^2+bn+c$ formunda. Baslangic degerleri $s(1)=4, s(2)=12,s(3)=24$ kullanirsak,
$ a+b+c=4$
$4 a+2 b+c=12 $
$9 a+3 b+c=24$
Sistemini elde ederiz. Cozum ise $a=2,b=2,c=0$, Yani $s(n)=2n^2+2n$ mis.
Bu da bize $$\sum_{n=1}^{20}1+\sum_{n=1}^{20}\dfrac{1}{2n^2+2n}=\sum_{n=1}^{20}1+\dfrac{1}{2}\sum_{n=1}^{20}\dfrac{1}{n(n+1)}$$
$$\dfrac{1}{n(n+1)}=\dfrac{A}{n}+\dfrac{B}{n+1}$$
$$1=A(n+1)+B(n)$$
$$1=A+(A+B)n\Longrightarrow A=1, B=-1$$
$$\sum_{n=1}^{20}1+\dfrac{1}{2}\sum_{n=1}^{20}\dfrac{1}{n(n+1)}=\sum_{n=1}^{20}1+\dfrac{1}{2}\sum_{n=1}^{20}\dfrac{1}{n}-\dfrac{1}{n+1}$$
$$20+\dfrac{1}{2}\Big(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{20}-\dfrac{1}{21}\Big)$$
$$20+\dfrac{1}{2}\Big(1-\dfrac{1}{21}\Big)=\dfrac{430}{21}$$