ページ3:

S=1² + 2 + + (n-1) + n = f(n+1) (2n+1)
(証明)
3
(k+1)² - k² = {(k+1) - k} {{k+1}'+ k (k+1) + k^}
-
41-3
3
=
3
11
=
=
=
k² + 2k+1+k² + k + k
3²+3+1
2
3.1 ± 3.1 + |
3.2
2
2
+
+3.2 +
1
= 3.3² + 3 3 + 1
+) (+)-4² = 3-4 +34 +1
(4+1)³-9³
3
3
2
(9 +1)³ - 1 ³ = 3 ( 1² + 2² + --- + r²)
2
+ 3 ( 1 + 2 + ...
+2
=
(9+1)
+ 1x2
=
3,5 + 3 x 1m (m+1) +m

ページ4:

3
3
35 = (n+1)³ - | - 2½ nim+1) - N
よって
2
= {(+1) - 1 } { (n+1)² + (+1) + 1} - 12 mm+1) - m
= n ( n² + 29+ | +n+1 +1 ) - 1/2 m (n+1)-m
= n ( n² + 3 ~ +3) - ≤ m (M+1)-~
=
1
!
±a (₂n² + bα + 6-30-3-2)
2
1
= 1
a (2 m² + 3 n + 1)
a (2n+1)(n+1)
Sn = f n (m+11 (2n+1)
(証明終)

ページ5:

2
√ = 1² + 2² +
(証明)
(n-1)+ é
- + = { }
n(n+1
(k+1)*-k" - {(k+1)² - k³{{(k+1)²+k"}
-14
=
=
=
319-29 =
(2k+1)(2k+2k+1)
4k²³ + 6k² + 4k+
4.13
2
1
+ 6.1² + 4·1
4.23 6.2
+
2
+ 4.2
+ 1
I ch+ist-m² = 4+m² + 6-m² + 4.h + 1
(n+1)9-1
=
=
3
4 ( 1³ + 2² + ....
2
6 (1² + 2 " +
+4(1+2+..
+(xr
+ n²)
+ n
り
m)
3
4 + 6 xα (n+1) (2n+1)
+4x+min+1)+om
=
=
+(9+1) (29+1)

ページ6:

4√ = (n+1)* -1 -n (n+1) (2n+1)-29(n+1)-n
= (n+1)+ -n (n+1) (2n+1)-29(n+1)-(~~~+1)
= (m+1) { (m+1)² - (2n+1)- 2m -1}
= (a+1) { (a+1) ³ - 2m² - n - 2m -1}
= (9+1) { (n+1)² - 2M (M +1)-(n+1))}
= (n+1)² { (n+1)² - 2m -1}
= (n+1)²n'
2
よって
=
=
acatio}"
IJ
(証明終)