Введение в математический анализ в вопросах и задачах. Анчиков А.М - 14 стр.

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8.
1
1·3
+
1
3·5
+ ... +
1
(2n1)(2n+1)
=
n
2n+1
;
9.
1
2
+ cos x + cos 2x + ... + cos nx =
sin
2n+1
2
x
2 sin
x
2
, x 6= 2πm;
10. cos x · cos 2x · ... · cos 2
n
x =
sin 2
n+1
x
2
n+1
·sin x
, x 6= πm;
11.
1
2
1·3
+
2
2
3·5
+ ... +
n
2
(2n1)(2n+1)
=
n(n+1)
2(2n+1)
;
12.
0
1!
+
1
2!
+
2
3!
+ ... +
n1
n!
= 1
1
n!
.
S
n
:
13. S
n
=
1
1·4
+
1
4·7
+
1
7·10
+ ... +
1
(3n2)(3n+1)
;
h
n
3n+1
i
.
14. S
n
= 1
x
1!
+
x(x1)
2!
...+(1)
n
x(x1)...(xn+1)
n!
;
h
(1)
n
(x1)...(xn)
n!
i
.
15. P
n
= (1
1
2
)(1
1
3
)(1
1
4
)...(1
1
n+1
);
h
1
1+n
i
.
16. P
n
= (1
1
4
)(1
1
9
)...(1
1
n
2
)
, n 2;
h
n+1
2n
i
.
17. 1 + x + x
2
+ ... + x
n
=
x
n+1
1
x1
, (x 6= 1);
18. 1·2·3+2·3·4+3·4·5 + ... + n(n + 1)(n+2) =
n(n+1)(n+2)(n+3)
4
;
19.
1
1+x
+
2
1+x
2
+
4
1+x
4
+ ... +
2
n
1+x
2
n
=
1
x1
+
2
n+1
1x
2
n+1
;
20. sin x + 2 sin 2x + 3 sin 3x + ... + n sin nx =
(n+1) sin nxn sin(n+1)x
4 sin
2
x
2
;
21. cos x+2 cos 2x+3 cos 3x+...+n cos nx =
(n+1) cos nxn cos(n+1)x1
4 sin
2
x
2
.
n
h
a
n
= a
1
+ (n 1)d, S
n
=
(a
1
+a
n
)n
2
; b
n
= b
1
q
n1
, S
n
= b
1
q
n
1
q1
i
.
       1            1                        1                    n
8.    1·3
               +   3·5
                         + ... +        (2n−1)(2n+1)
                                                         =      2n+1
                                                                     ;
      1                                                                   sin 2n+1   x
9.    2
          + cos x + cos 2x + ... + cos nx =                                      2
                                                                            2 sin x2
                                                                                       ,   x 6= 2πm;
                                                               sin 2n+1 x
10. cos x · cos 2x · ... · cos 2n x =                          2n+1 ·sin x
                                                                           ,      x 6= πm;
          12        22                         n2                 n(n+1)
11.       1·3
                +   3·5
                             + ... +      (2n−1)(2n+1)
                                                             =    2(2n+1)
                                                                          ;
          0        1         2                 n−1               1
12.       1!
               +   2!
                         +   3!
                                  + ... +       n!
                                                     =1−         n!
                                                                    .
Íàéòè çíà÷åíèÿ ñóìì Sn :
                                                                                               h             i
                        1          1          1                    1                                 n
13. Sn =               1·4
                             +    4·7
                                        +   7·10
                                                   + ... +    (3n−2)(3n+1)
                                                                           ;                       3n+1
                                                                                                                 .
                                                                                               h                      i
14. Sn = 1− 1!x + x(x−1)
                    2!
                         −...+(−1)n x(x−1)...(x−n+1)
                                           n!
                                                     ; (−1)n (x−1)...(x−n)
                                                                  n!
                                                                           .
Íàéòè çíà÷åíèå âûðàæåíèé:
                                                                                           h         i
15. Pn = (1 − 12 )(1 − 31 )(1 − 14 )...(1 −                                1
                                                                          n+1
                                                                              );                1
                                                                                               1+n
                                                                                                         .
                                                                                           h         i
16. Pn = (1 − 14 )(1 − 91 )...(1 −                            1
                                                             n2 )
                                                                  ,     n ≥ 2;                 n+1
                                                                                                2n
                                                                                                         .
Äîêàçàòü:
                                                      xn+1 −1
17. 1 + x + x2 + ... + xn =                             x−1
                                                              ,         (x 6= 1);
                                                                                                     n(n+1)(n+2)(n+3)
18. 1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + ... + n(n + 1)(n + 2) =                                              4
                                                                                                                      ;
           1              2               4                2n              1            2n+1
19.       1+x
                   +    1+x2
                                  +     1+x4
                                               + ... +   1+x2n
                                                                      =   x−1
                                                                                 +    1−x2
                                                                                           n+1 ;

                                                                                           (n+1) sin nx−n sin(n+1)x
20. sin x + 2 sin 2x + 3 sin 3x + ... + n sin nx =                                                  4 sin2 x2
                                                                                                                    ;
                                                                                     (n+1) cos nx−n cos(n+1)x−1
21. cos x+2 cos 2x+3 cos 3x+...+n cos nx =                                                      4 sin2 x2
                                                                                                                .
Ïðèìåíÿÿ ìåòîä ìàòåìàòè÷åñêîé èíäóêöèè:
22. Âûðàçèòü n-é ÷ëåí àðèôìåòè÷åñêîé, ãåîìåòðè÷åñêîé ïðî-
ãðåññèé ÷åðåç ïåðâûé ÷ëåí ñîîòâåòñòâåííî. Íàéòè çíà÷åíèÿ ñóìì
ýòèõ
h
      ïðîãðåññèé.                                                          i
                                                                    q n −1
  an = a1 + (n − 1)d, Sn = (a1 +a
                                2
                                  n )n
                                       ; b n = b 1 q n−1
                                                         , S n = b1  q−1
                                                                             .

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