Введение в математический анализ в вопросах и задачах. Анчиков А.М - 16 стр.

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a) C
x2
x+1
+ 2C
3
x1
= 7(x 1); [x = 5] .
b) C
1
x
+ 6C
2
x
+ 6C
3
x
= 9x
2
14; [x = 7] .
(2x +
1
x
2
)
m
x.
x
(1 + x
3
)
30
? [ x = 2 ] .
|x
1
+ x
2
+ ... + x
n
| |x
1
| + |x
2
| + ... + |x
n
|.
2! · 4! · 6! · ... · (2n)! > [(n + 1)!]
n
; n 2.
2
n
> 2n + 1; n 3.
32. (0, 7)
n
1 0, 3n.
33. (1 + x
1
)(1 + x
2
)...(1 + x
n
) 1 + x
1
+ x
2
+ ... + x
n
,
x
1
, x
2
, ..., x
n
34. a) 2
n
> n
2
, n = 1, n 5; b) 3
n
> n
3
, n 4, n = 1, 2.
35.
n < 1 +
1
2
+
1
3
+ ... +
1
n
< 2
n, n 2.
1
1+k
>
k + 1
k
36. 2
n(n1)
2
> n!, n 3.
37. n
n
2
< n! < (
n+1
2
)
n
, n 2.
38.
2
2n
n+1
<
(2n)!
(n!)
2
, n 2 .
39. n
n+1
> (n + 1)
n
, n 3.
40. (2n)! < 2
2n
(n!)
2
.
27. Ðåøèòü óðàâíåíèÿ:
    x−2     3
a) Cx+1 + 2Cx−1 = 7(x − 1);             [x = 5] .

b) Cx1 + 6Cx2 + 6Cx3 = 9x2 − 14; [x = 7] .
28. Òðåòüå ñëàãàåìîå ðàçëîæåíèÿ (2x + x12 )m íå ñîäåðæèò x.
Ïðè êàêèõ çíà÷åíèÿõ x ýòî ñëàãàåìîå ðàâíî âòîðîìó ñëàãàå-
ìîìó ðàçëîæåíèÿ (1 + x3 )30 ? [ x = 2 ] .
Äîêàçàòü:
29. |x1 + x2 + ... + xn | ≤ |x1 | + |x2 | + ... + |xn | .
30. 2! · 4! · 6! · ... · (2n)! > [(n + 1)!]n ; ∀n ≥ 2.
31. 2n > 2n + 1; ∀n ≥ 3.
32. (0, 7)n ≥ 1 − 0, 3n.
33. (1 + x1 )(1 + x2 )...(1 + xn ) ≥ 1 + x1 + x2 + ... + xn ,
ãäå x1 , x2 , ..., xn − ÷èñëà îäíîãî çíàêà, áîëüøèå -1 (íåðàâåí-
ñòâî Áåðíóëëè).
34. a) 2n > n2 , n = 1, ∀n ≥ 5; b) 3n > n3 , n ≥ 4, n = 1, 2.
    √                                √
35. n < 1 + √12 + √13 + ... + √1n < 2 n, ∀n ≥ 2.
                                    1
                                         √         √
[ Âîñïîëüçîâàòüñÿ íåðàâåíñòâîì √1+k    > k + 1 − k ].
        n(n−1)
36. 2      2     > n!, ∀n ≥ 3.
        n
37. n 2 < n! < ( n+1
                  2
                     )n , ∀n ≥ 2.
      22n        (2n)!
38.   n+1
            <    (n!)2
                       , ∀n   ≥ 2.
39. nn+1 > (n + 1)n , ∀n ≥ 3.
40. (2n)! < 22n (n!)2 .

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