Using the sum of general term, 1 + 11 + 111 + +111… The sum of 1n ones

Using the sum of general term, 1 + 11 + 111 + +111… The sum of 1n ones

Because of 111 1n 1 = 19 × 999 9N = 10n-19 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 1 + 11 + 111 + +111… 1n 1 = 19 [(10-1) + (102-1) + +(10n-1)]=19(10+102+… +10n)-n9=19•10(1-10n)1-10-n9=10n+1-9n-1081
Sequence summation Sn = 1 + 11 + 111 + 1111 +... + 1111... 111 (n 1)
Summation of sequence
Sn = 1 + 11 + 111 + 1111 +... + 1111... 111 (n 1)
How can this kind of thing be extended from Sn = 9 + 99 + 999 +... + 9999.. 9 (n 9S)?
Sn = 1 + 11 + 111 + 1111 +... + 1111... 111 (n 1)
an=(1/9)*(10^n-1)
SN=(1/9)*(10-10^(n+1))/(1-10)-n=(10/81)(10^n-1)-n
9 * SN1 = SN2, and SN1 + SN2 = 11 10 (n + 1, 1, 0). So SN1 = 1111 1 (n + 1)
An = 1 / 9 (10 ^ n-1)
The latter is nine times the former
It is proved that CO α t-tan α = (2cos ^ 2 α - 1) / sin α * cos α
coαt-tanα=cosa/sina-sina/cosa=cos^2a-sin^2a/sinα*cosα
=cos^2a-(1-cos^2a)/sinα*cosα
=（2cos^2α-1）/sinα*cosα
Given the function f (x) = log (4 ^ x + 1) - ax 1 with base 2, if the function is an even function on R, find the value of real number a
1. If the function is an even function on R, find the value of real number a
2. If a = 4, find the zero point of function f (x)
The function is an even function on R. if the derivative of F (x) at x = 0 is 0, it is easy to calculate a = 1 / (2ln2). F (x) = f (- x), i.e. log (4 ^ x + 1) - AX = log [4 ^ (- x) + 1] + ax, log (4 ^ x + 1) - log [4 ^ (- x) + 1] = 2aX, i.e. L
If f (x) is an even function on R, find the value of real number A. if a = 4, find the zero point 1 of function f (x). If f is negative x equals FX, the left side is 2aX, and the right side is log2. The bottom exponent is fraction. The numerator is 4 times x plus 1, and the denominator is 4 times negative x plus 1. The fraction is approximately 4 The whole right side of X is 2x, so a is 1 2
If f (x) is an even function on R, find the value of real number A. if a = 4, find the zero point of function f (x)
Given that f (x) = e ^ x + X, G (x) = LNX + X, H (x) = lnx-1, the zeros are a, B, C in turn. Try to judge the size of a, B, C
e^a+a=0,a=-e^a b>0,lnb=-b 0
In the known sequence an, A1 = 1, when n ≥ 2, the sum of the first n terms is Sn, satisfying Sn & # 178; = an (sn-1)
1. Prove that 1 / Sn is an arithmetic sequence 2. Prove the first n terms and TN of BN = log2 Sn / Sn + 2, {BN}, and find the smallest positive integer satisfying t ≥ 6
Finding the smallest positive integer n satisfying TN greater than or equal to 6
Let me give you a detailed answer
（1）Sn²=an(Sn-1)
Sn²=[sn-s(n-1)]*（sn-1)
=Sn²-sn*sn(n-1)-sn+sn(n-1)
Sn-Sn (n-1) = - Sn * Sn (n-1) divided by SN * Sn (n-1)
1/sn-1/sn(n-1)=1 （n.>=2,n∈N*）
After testing, when = 1, the original formula still holds
（2）Bn=Log2 （sn/sn+2）
Using the logarithmic expansion formula (the ratio moved outside is subtraction)
Bn=Log2（ sn/sn+2）=,Bn=Log2（ sn）-Log2（sn+2）
Let's start with B1, B1 = log2 (S1) - log2 (S3)
B2=Log2（ s2）-Log2（s4）
B3= Log2（ s3）-Log2（s5）
B4=Log2（ s4）-Log2（s6）…… You have found the regularity and repeating elements. These elements added to BN are TN, and the positive and negative ones in the middle cancel each other, leaving the four terms log2 (S1) + log2 (S2) - log2 (Sn + 1) - log2 (Sn + 2) (you can list the following items, and you can find that the two items are subtracted from the latter, and the former does not cancel each other like him). TN = log2 (S1) + log2 (S2) - log2 (Sn + 1) - log2 (Sn + 2), put them together
=log2[（S1*s2)/(sn+1)*(sn+2）]
In addition, Sn = 1 / N (n ∈ n *)
Tn=Log2[（1/2）*(n+1)(n+2)]>=6
Both sides take 2 as the bottom power, and the sixth power of (1 / 2) * (n + 1) (n + 2) > = 2 = 64
(n+1)(n+2）>=128
We know that 11 & # 178; = 121
11 * 12 = 132 is more than 10 * 11 = 110, N + 1 = 10
N = 9 should be the largest positive integer, 9 is not the smallest positive integer
Sn²=an(Sn-1)
Sn²=[sn-s(n-1)]*（sn-1)=Sn²-sn*sn(n-1)-sn+sn(n-1)
Sn-Sn (n-1) = - Sn * Sn (n-1) divided by SN * Sn (n-1)
1 / sn-1 / Sn (n-1) = 1 sequence Sn, 1 / 2 is arithmetic sequence
2. The general term formula of ball sequence an
1 / Sn... Expand
Sn²=an(Sn-1)
Sn²=[sn-s(n-1)]*（sn-1)=Sn²-sn*sn(n-1)-sn+sn(n-1)
Sn-Sn (n-1) = - Sn * Sn (n-1) divided by SN * Sn (n-1)
1 / sn-1 / Sn (n-1) = 1 sequence Sn, 1 / 2 is arithmetic sequence
2. The general term formula of ball sequence an
1/sn-1/sn(n-1)=1
1/（sn-1）-1/sn(n-2)=1
:
1 / s2-1 / S1 = 1
1/sn-1/s1=n-1
1/sn=n
sn=1/n
an=sn-sn-1=1/n-1/(n-1）=-1/n(n-1)
[or use Sn & # 178; = an (sn-1) an = Sn & # 178; / (sn-1) = 1 / N & # 178; / (1-N) / N] = - 1 / N (n-1)]
[n = 1, an = 1, when n ≥ 2, an = - 1 / N (n-1)]
Prove 1-2cos α squared / sin α cos α = Tan α - 1 / Tan α
1-2cos α squared = (sin α) ^ 2 + (COS α) ^ 2-2 (COS α) ^ 2 = = (sin α) ^ 2 - (COS α) ^ 2 1-2cos α squared / sin α cos α = Tan α - 1 / Tan α
The mathematics of the first year of senior high school should have a detailed process, f (x) = log, with a as the base (x + 8-A / x) is an increasing function on [1, + ∞), to find the value range of the real number a
F (x) = log with a as the base (x + 8-A / x) is an increasing function on [1, + ∞). Find the value range of real number a
Because the domain [1, + ∞] of F (x) is defined by a function
X + 8-A / X must be greater than 0, so there is A1 with 1 + 8-A > 0
To sum up, the value range of a is 1
F (x) = log with a as the base (x + 8-A / x) is an increasing function on [1, + ∞)
Then, x + 8-A / x > 0
A>1
That is: (x ^ 2 + 8x-a) x > 0
A>1
If the inequality has a real solution to x, the range of a can be obtained
Given the function f (x) = (e ^ x) / (x ^ 2-ax + 1) 1. Find the monotone interval 2. If the inequality f (x) is greater than or equal to x, it belongs to [0, a + 1] for any X
Solution 1
f(x)=(e^x)/(x²-ax+1)
f'(x)=[(e^x)'(x²-ax+1)-(e^x)(x²-ax+1)']/(x²-ax+1)²
f'(x)=[(e^x)(x²-ax+1)-(e^x)(2x-a)]/(x²-ax+1)²
f'(x)=[(e^x)(x²-ax+1-2x+a)]/(x²-ax+1)²
f'(x)=(e^x)[x²-(a+2)x+a+1]/(x²-ax+1)²
1. Let f '(x) > 0, that is: (e ^ x) [x & # 178; - (a + 2) x + A + 1] / (X & # 178; - ax + 1) &# 178; > 0
There are: X & # 178; - (a + 2) x + A + 1 > 0
x²-2×[(a+2)/2]x+[(a+2)/2]²-[(a+2)/2]²+a+1＞0
[x-(a+2)/2]²-[(a+2)/2]²+a+1＞0
[x-(a+2)/2]²-[(a²+4a+4-4a-4)/4＞0
[x-(a+2)/2]²＞(a/2)²
When a ≥ 0:
x-(a+2)/2＞a/2、x-(a+2)/2＜-a/2
The solution is: x > A + 1, x < 1
When a < 0:
x-(a+2)/2＞-a/2、x-(a+2)/2＜a/2
The solution is: x > 1, x < A + 1
2. Let: F '(x) < 0, that is: (e ^ x) [x & # 178; - (a + 2) x + A + 1] / (X & # 178; - ax + 1) &# 178; < 0
There are: X & # 178; - (a + 2) x + A + 1 < 0
x²-2×[(a+2)/2]x+[(a+2)/2]²-[(a+2)/2]²+a+1＜0
[x-(a+2)/2]²-[(a+2)/2]²+a+1＜0
[x-(a+2)/2]²-[(a²+4a+4-4a-4)/4＜0
[x-(a+2)/2]²＜(a/2)²
When a ≥ 0:
-a/2＜x-(a+2)/2＜a/2
The solution is: 1 < x < A + 1
When a < 0:
a/2＜x-(a+2)/2＜-a/2
The solution is a + 1 < x < 1
To sum up, the interval is monotonic
When a ≥ 0:
The monotone increasing interval of F (x) is: X ∈ (- ∞, 1), X ∈ (a + 1, ∞)
The monotone decreasing interval of F (x) is: X ∈ (1, a + 1)
When a < 0:
The monotone increasing interval of F (x) is: X ∈ (- ∞, a + 1), X ∈ (1, ∞)
The monotone decreasing interval of F (x) is: X ∈ (a + 1,1)
Solution 1
f(x)=(e^x)/(x²-ax+1)
f'(x)=[(e^x)'(x²-ax+1)-(e^x)(x²-ax+1)']/(x²-ax+1)²
f'(x)=[(e^x)(x²-ax+1)-(e^x)(2x-a)]/(x²-ax+1)²
F '(x) = [(e ^ x) (X & # 178; - a... expand
Solution 1
f(x)=(e^x)/(x²-ax+1)
f'(x)=[(e^x)'(x²-ax+1)-(e^x)(x²-ax+1)']/(x²-ax+1)²
f'(x)=[(e^x)(x²-ax+1)-(e^x)(2x-a)]/(x²-ax+1)²
f'(x)=[(e^x)(x²-ax+1-2x+a)]/(x²-ax+1)²
f'(x)=(e^x)[x²-(a+2)x+a+1]/(x²-ax+1)²
1. Let f '(x) > 0, that is: (e ^ x) [x & # 178; - (a + 2) x + A + 1] / (X & # 178; - ax + 1) &# 178; > 0
There are: X & # 178; - (a + 2) x + A + 1 > 0
x²-2×[(a+2)/2]x+[(a+2)/2]²-[(a+2)/2]²+a+1＞0
[x-(a+2)/2]²-[(a+2)/2]²+a+1＞0
[x-(a+2)/2]²-[(a²+4a+4-4a-4)/4＞0
[x-(a+2)/2]²＞(a/2)²
When a ≥ 0:
x-(a+2)/2＞a/2、x-(a+2)/2＜-a/2
The solution is: x > A + 1, x < 1
When a < 0:
x-(a+2)/2＞-a/2、x-(a+2)/2＜a/2
The solution is: x > 1, x < A + 1
2. Let: F '(x) < 0, that is: (e ^ x) [x & # 178; - (a + 2) x + A + 1] / (X & # 178; - ax + 1) &# 178; < 0
There are: X & # 178; - (a + 2) x + A + 1 < 0
x²-2×[(a+2)/2]x+[(a+2)/2]²-[(a+2)/2]²+a+1＜0
[x-(a+2)/2]²-[(a+2)/2]²+a+1＜0
[x-(a+2)/2]²-[(a²+4a+4-4a-4)/4＜0
[x-(a+2)/2]²＜(a/2)²
When a ≥ 0:
-a/2＜x-(a+2)/2＜a/2
The solution is: 1 < x < A + 1
When a < 0:
a/2＜x-(a+2)/2＜-a/2
The solution is a + 1 < x < 1
In conclusion, the monotone interval is as follows:
When a ≥ 0:
The monotone increasing interval of F (x) is: X ∈ (- ∞, 1), X ∈ (a + 1, ∞)
The monotone decreasing interval of F (x) is: X ∈ (1, a + 1)
When a < 0:
The monotone increasing interval of F (x) is: X ∈ (- ∞, a + 1), X ∈ (1, ∞)
The monotone decreasing interval of F (x) is: X ∈ (a + 1,1). Put it away
It is known that the sequence an is an equal ratio sequence with the first term sum of 1 and the common ratio of 2. The first n terms of BN and Sn = n ^ 2
1. Finding the general term formula of [an] and [BN]
2. Finding the sum of the first n terms of the sequence [BN * an]
1. When the general term formula of an = 1 × 2 ^ (n-1) = 2 ^ (n-1) sequence {an} is an = 2 ^ (n-1) n = 1, B1 = S1 = 1 & # 178; = 1n ≥ 2, Sn = n & # 178; s (n-1) = (n-1) &# 178; BN = SN-S (n-1) = n & # 178; - (n-1) &# 178; = 2n-1n = 1, B1 = 2-1 = 1. The general term formula of sequence {BN} is BN = 2n-12