1
∗
2
+
2
∗
3
+
3
∗
4
+
⋯
+
n
(
n
+
1
)
=
n
(
n
+
1
)
(
n
+
2
)
3
{\displaystyle 1*2+2*3+3*4+\cdots +n(n+1)={\frac {n(n+1)(n+2)}{3}}}
n
=
1
{\displaystyle \ n=1}
עריכה
L
:
n
(
n
+
1
)
=
1
∗
2
=
2
R
:
n
(
n
+
1
)
(
n
+
2
)
3
=
6
3
=
2
2
=
2
{\displaystyle {\begin{aligned}&L:n(n+1)=1*2=2\\&R:{\frac {n(n+1)(n+2)}{3}}={\frac {6}{3}}=2\\&2=2\\\end{aligned}}}
n
=
k
{\displaystyle \ n=k}
עריכה
1
∗
2
+
2
∗
3
+
3
∗
4
+
⋯
+
k
(
k
+
1
)
=
k
(
k
+
1
)
(
k
+
2
)
3
{\displaystyle 1*2+2*3+3*4+\cdots +k(k+1)={\frac {k(k+1)(k+2)}{3}}}
עריכה
1
∗
2
+
2
∗
3
+
3
∗
4
+
⋯
+
k
(
k
+
1
)
⏟
=
k
(
k
+
1
)
(
k
+
2
)
3
+
(
k
+
1
)
(
k
+
2
)
=
(
k
+
1
)
(
k
+
2
)
(
k
+
3
)
3
k
(
k
+
1
)
(
k
+
2
)
3
+
(
k
+
1
)
(
k
+
2
)
=
(
k
+
1
)
(
k
+
2
)
(
k
+
3
)
3
k
(
k
+
1
)
(
k
+
2
)
+
3
(
k
+
1
)
(
k
+
2
)
=
(
k
+
1
)
(
k
+
2
)
(
k
+
3
)
/
:
(
k
+
1
)
(
k
+
2
)
k
+
3
=
k
+
3
{\displaystyle {\begin{aligned}&\underbrace {1*2+2*3+3*4+\cdots +k(k+1)} _{={\frac {k(k+1)(k+2)}{3}}}+(k+1)(k+2)={\frac {(k+1)(k+2)(k+3)}{3}}\\&{\frac {k(k+1)(k+2)}{3}}+(k+1)(k+2)={\frac {(k+1)(k+2)(k+3)}{3}}\\&k(k+1)(k+2)+3(k+1)(k+2)=(k+1)(k+2)(k+3)/:(k+1)(k+2)\\&k+3=k+3\\\end{aligned}}}
הסדרה :
(
n
+
1
)
(
n
+
2
)
+
(
n
+
2
)
(
n
+
3
)
+
(
n
+
3
)
(
n
+
4
)
+
⋯
+
2
n
(
2
n
+
1
)
{\displaystyle (n+1)(n+2)+(n+2)(n+3)+(n+3)(n+4)+\cdots +2n(2n+1)}
1
∗
2
+
2
∗
3
+
3
∗
4
+
⋯
+
n
(
n
+
1
)
=
n
(
n
+
1
)
(
n
+
2
)
3
{\displaystyle 1*2+2*3+3*4+\cdots +n(n+1)={\frac {n(n+1)(n+2)}{3}}}
a
2
n
=
(
n
+
1
)
(
n
+
2
)
+
(
n
+
2
)
(
n
+
3
)
+
(
n
+
3
)
(
n
+
4
)
+
⋯
+
2
n
(
2
n
+
1
)
{\displaystyle a_{2n}=(n+1)(n+2)+(n+2)(n+3)+(n+3)(n+4)+\cdots +2n(2n+1)}
הסדרה המלא :
a
n
=
1
∗
2
+
2
∗
3
+
3
∗
4
+
⋯
+
n
(
n
+
1
)
+
(
n
+
1
)
(
n
+
2
)
+
(
n
+
2
)
(
n
+
3
)
+
(
n
+
3
)
(
n
+
4
)
+
⋯
+
2
n
(
2
n
+
1
)
a
1
n
=
1
∗
2
+
2
∗
3
+
3
∗
4
+
⋯
+
n
(
n
+
1
)
=
n
(
n
+
1
)
(
n
+
2
)
3
a
2
n
=
(
n
+
1
)
(
n
+
2
)
+
(
n
+
2
)
(
n
+
3
)
+
(
n
+
3
)
(
n
+
4
)
+
⋯
+
2
n
(
2
n
+
1
)
{\displaystyle {\begin{aligned}&an=1*2+2*3+3*4+\cdots +n(n+1)+{\color {green}(n+1)(n+2)+(n+2)(n+3)+(n+3)(n+4)+\cdots +2n(2n+1)}\\&a_{1n}=1*2+2*3+3*4+\cdots +n(n+1)={\frac {n(n+1)(n+2)}{3}}\\&a_{2n}=(n+1)(n+2)+(n+2)(n+3)+(n+3)(n+4)+\cdots +2n(2n+1)\\\end{aligned}}}
S
a
n
−
S
1
a
n
=
S
2
a
n
{\displaystyle \ S_{an}-S_{1an}=S_{2an}}
1
+
2
+
3
+
4
+
5
+
6
+
7
=
?
{\displaystyle \ 1+2+3+4+5+6+7=?}
נרשום את הנאמר בשפת המתמטיקה :
a
n
−
a
1
n
=
1
∗
2
+
2
∗
3
+
3
∗
4
+
⋯
+
n
(
n
+
1
)
+
(
n
+
1
)
(
n
+
2
)
+
(
n
+
2
)
(
n
+
3
)
+
(
n
+
3
)
(
n
+
4
)
+
⋯
+
2
n
(
2
n
+
1
)
−
1
∗
2
+
2
∗
3
+
3
∗
4
+
⋯
+
n
(
n
+
1
)
{\displaystyle an-a_{1n}={\color {green}1*2+2*3+3*4+\cdots +n(n+1)+{\color {green}(n+1)(n+2)+(n+2)(n+3)+(n+3)(n+4)+\cdots +2n(2n+1)}}-1*2+2*3+3*4+\cdots +n(n+1)}
אנו יודעים (על פי תרגיל א') שסכום הסדרה הראשונה שווה ל :
1
∗
2
+
2
∗
3
+
3
∗
4
+
⋯
+
n
(
n
+
1
)
=
n
(
n
+
1
)
(
n
+
2
)
3
{\displaystyle 1*2+2*3+3*4+\cdots +n(n+1)={\color {blue}{\frac {n(n+1)(n+2)}{3}}}}
אנו רוצים למצוא את סכום כל הסדרה כולל (
a
n
{\displaystyle \ an}
a
n
{\displaystyle \ an}
a
1
n
{\displaystyle \ a_{1}n}
(
n
(
n
+
1
)
(
n
+
2
)
3
{\displaystyle {\frac {n(n+1)(n+2)}{3}}}
שכן, מדובר באותה סדרה ולפיכך, אופן חישוב הסכום זהה. הפעולה אולי נשמעת קשה, אך, היא פשוטה למדי, כל שעלינו לעשות הוא להציב 2n במקום 1n (n).
סכום הסדרה הכוללת הוא :
S
2
a
n
=
2
n
(
2
n
+
1
)
(
2
n
+
2
)
3
{\displaystyle S_{2an}=\color {green}{\frac {2n(2n+1)(2n+2)}{3}}}
נציב את סכומי הסדרות :
2
n
(
2
n
+
1
)
(
2
n
+
2
)
3
−
n
(
n
+
1
)
(
n
+
2
)
3
{\displaystyle {\color {green}{\frac {2n(2n+1)(2n+2)}{3}}}-{\color {blue}{\frac {n(n+1)(n+2)}{3}}}}
ועתה נשאר רק לפתור :
2
n
(
2
n
+
1
)
(
2
n
+
2
)
3
−
n
(
n
+
1
)
(
n
+
2
)
3
2
n
(
2
n
+
1
)
(
2
n
+
2
)
−
n
(
n
+
1
)
(
n
+
2
)
3
2
n
(
2
n
+
1
)
2
(
n
+
1
)
−
n
(
n
+
1
)
(
n
+
2
)
3
[
n
(
n
+
1
)
]
[
4
(
2
n
+
1
)
−
(
n
+
2
)
]
3
n
(
n
+
1
)
(
8
n
+
4
−
n
−
2
)
3
n
(
n
+
1
)
(
7
n
−
2
)
3
{\displaystyle {\begin{aligned}&{\color {green}{\frac {2n(2n+1)(2n+2)}{3}}}-{\color {blue}{\frac {n(n+1)(n+2)}{3}}}\\&{\frac {2n(2n+1)(2n+2)-n(n+1)(n+2)}{3}}\\&{\frac {2n(2n+1)2(n+1)-n(n+1)(n+2)}{3}}\\&{\frac {[n(n+1)][4(2n+1)-(n+2)]}{3}}\\&{\frac {n(n+1)(8n+4-n-2)}{3}}\\&{\frac {n(n+1)(7n-2)}{3}}\\\end{aligned}}}
(
n
+
1
)
(
n
+
2
)
+
(
n
+
2
)
(
n
+
3
)
+
(
n
+
3
)
(
n
+
4
)
+
⋯
+
2
n
(
2
n
+
1
)
=
n
(
n
+
1
)
(
7
n
−
2
)
3
{\displaystyle (n+1)(n+2)+(n+2)(n+3)+(n+3)(n+4)+\cdots +2n(2n+1)={\frac {n(n+1)(7n-2)}{3}}}
n
=
1
{\displaystyle \ n=1}
עריכה
L
:
2
n
(
2
n
+
1
)
=
2
∗
1
(
2
∗
1
+
1
)
→
(
n
+
1
)
(
n
+
2
)
=
6
R
:
n
(
n
+
1
)
(
7
n
−
2
)
3
=
6
6
=
6
{\displaystyle {\begin{aligned}&L:2n(2n+1)=2*1(2*1+1)\rightarrow (n+1)(n+2)=6\\&R:{\frac {n(n+1)(7n-2)}{3}}=6\\&6=6\\\end{aligned}}}
n
=
k
{\displaystyle \ n=k}
עריכה
(
k
+
1
)
(
k
+
2
)
+
(
k
+
2
)
(
k
+
3
)
+
(
k
+
3
)
(
k
+
4
)
+
⋯
+
2
k
(
2
k
+
1
)
=
k
(
k
+
1
)
(
7
k
+
2
)
3
{\displaystyle (k+1)(k+2)+(k+2)(k+3)+(k+3)(k+4)+\cdots +2k(2k+1)={\frac {k(k+1)(7k+2)}{3}}}
עריכה
(
k
+
2
)
(
k
+
3
)
+
(
k
+
3
)
(
k
+
4
)
+
⋯
+
2
k
(
2
k
+
1
)
⏟
=
k
(
k
+
1
)
(
7
k
−
2
)
3
−
(
k
+
10
(
k
+
2
)
+
(
2
k
+
1
)
(
2
k
+
2
)
+
(
2
k
+
2
)
(
2
k
+
3
)
=
(
k
+
1
)
(
k
+
2
)
(
7
k
+
9
)
3
k
(
k
+
1
)
(
7
k
−
2
)
3
−
(
k
+
1
)
(
k
+
2
)
+
(
2
k
+
1
)
(
2
k
+
2
)
+
(
2
k
+
2
)
(
2
k
+
3
)
=
(
k
+
1
)
(
k
+
2
)
(
7
k
+
9
)
3
k
(
k
+
1
)
(
7
k
+
2
)
−
3
(
k
+
1
)
(
k
+
2
)
+
6
(
2
k
+
1
)
(
k
+
1
)
+
6
(
k
+
1
)
(
2
k
+
3
)
=
(
k
+
1
)
(
k
+
2
)
(
7
k
+
9
)
/
:
(
k
+
1
)
k
(
7
k
+
2
)
−
3
(
k
+
2
)
+
6
(
2
k
+
1
)
+
6
(
2
k
+
3
)
=
(
k
+
2
)
(
7
k
+
9
)
7
k
2
+
2
k
−
3
k
−
6
+
12
k
+
6
+
12
k
+
18
=
7
k
2
+
9
k
+
14
k
+
18
7
k
2
+
23
k
+
18
=
7
k
2
+
23
k
+
18
0
=
0
{\displaystyle {\begin{aligned}&\underbrace {(k+2)(k+3)+(k+3)(k+4)+\cdots +2k(2k+1)} _{={\frac {k(k+1)(7k-2)}{3}}-(k+10(k+2)}+(2k+1)(2k+2)+(2k+2)(2k+3)={\frac {(k+1)(k+2)(7k+9)}{3}}\\&{\frac {k(k+1)(7k-2)}{3}}-(k+1)(k+2)+(2k+1)(2k+2)+(2k+2)(2k+3)={\frac {(k+1)(k+2)(7k+9)}{3}}\\&k(k+1)(7k+2)-3(k+1)(k+2)+6(2k+1)(k+1)+6(k+1)(2k+3)=(k+1)(k+2)(7k+9)/:(k+1)\\&k(7k+2)-3(k+2)+6(2k+1)+6(2k+3)=(k+2)(7k+9)\\&7k^{2}+2k-3k-6+12k+6+12k+18=7k^{2}+9k+14k+18\\&7k^{2}+23k+18=7k^{2}+23k+18\\&0=0\\&\end{aligned}}}
![{\displaystyle {\begin{aligned}&{\color {green}{\frac {2n(2n+1)(2n+2)}{3}}}-{\color {blue}{\frac {n(n+1)(n+2)}{3}}}\\&{\frac {2n(2n+1)(2n+2)-n(n+1)(n+2)}{3}}\\&{\frac {2n(2n+1)2(n+1)-n(n+1)(n+2)}{3}}\\&{\frac {[n(n+1)][4(2n+1)-(n+2)]}{3}}\\&{\frac {n(n+1)(8n+4-n-2)}{3}}\\&{\frac {n(n+1)(7n-2)}{3}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/346133022dbe20f655bfbab19e37b717c69a6e36)
