Введение в математический анализ в вопросах и задачах. Анчиков А.М - 15 стр.

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n k
C
k
n
=
n(n1)(n2)...(nk+1)
k!
=
n!
k!(nk)!
n k.
C
1
n
= n; C
n
n
= C
0
n
= 1; 0! = 1.
a) C
k
n
= C
nk
n
;
b) C
n
n+1
= C
k
n
+ C
k1
n
;
c) C
m
n
C
mk
nk
= C
k
m
C
m
n
;
d) C
k
n
+ 3C
k1
n
+ 3C
k2
n
+ C
k3
n
= C
k
n+3
;
e)
C
1
n
+ 2C
2
n
+ ... + nC
n
n
n
= 2
n1
.
(a + b)
n
= C
0
n
a
n
b
0
+ C
1
n
a
n1
b + ... + C
k
n
a
nk
b
k
+ ... + C
n
n
a
0
b
n
. (19)
a) C
0
n
+ C
1
n
+ ... + C
n
n
= 2
n
; ;
b) C
0
n
C
1
n
+ C
2
n
C
3
n
+ ... + (1)
n
C
n
n
= 0.
(a
1
+ a
2
+ ... + a
s
)
2
=
s
X
i=1
a
2
i
+ 2
X
i<j
a
i
a
j
. (20)
(
s
P
i=1
a
i
)
n
.
23. Ñî÷åòàíèÿìè èç n ýëåìåíòîâ ïî k íàçûâàþòñÿ òàêèå
ñîåäèíåíèÿ, êîòîðûå îòëè÷àþòñÿ äðóã îò äðóãà òîëüêî ñâîè-
ìè ýëåìåíòàìè. Cnk = n(n−1)(n−2)...(n−k+1)
                                k!
                                                n!
                                           = k!(n−k)! − ÷èñëî
ñî÷åòàíèé èç n ýëåìåíòîâ ïî k. Ïîëüçóÿñü ýòîé ôîðìó-
ëîé, ëåãêî óáåäèòüñÿ â ñïðàâåäëèâîñòè ñëåäóþùèõ ðàâåíñòâ:
Cn1 = n; Cnn = Cn0 = 1; 0! = 1.
Äîêàçàòü òîæäåñòâà:
a) Cnk = Cnn−k ;
    n
b) Cn+1 = Cnk + Cnk−1 ;
        m−k
c) Cnm Cn−k    k m
            = Cm Cn ;
                                    k
d) Cnk + 3Cnk−1 + 3Cnk−2 + Cnk−3 = Cn+3 ;
    Cn1 + 2Cn2 + ... + nCnn
e)                          = 2n−1 .
               n
24. Ïðèìåíÿÿ ìåòîä ìàòåìàòè÷åñêîé èíäóêöèè, ïîëó÷èòü ôîð-
ìóëó áèíîìà Íüþòîíà:

 (a + b)n = Cn0 an b0 + Cn1 an−1 b + ... + Cnk an−k bk + ... + Cnn a0 bn . (19)

25. Äîêàçàòü òîæäåñòâà:
a) Cn0 + Cn1 + ... + Cnn = 2n ; ;
b) Cn0 − Cn1 + Cn2 − Cn3 + ... + (−1)n Cnn = 0.
26. Äîêàçàòü ôîðìóëó:
                                          s
                                          X               X
               (a1 + a2 + ... + as )2 =         a2i + 2         ai aj .   (20)
                                          i=1             i