The sum of 1 / 2 + 5 / 6 + 11 / 12 + 19 / 20 + 29 / 30 +, 9701 / 9702 + 9899 / 9900 is calculated by recurrence equation

The sum of 1 / 2 + 5 / 6 + 11 / 12 + 19 / 20 + 29 / 30 +, 9701 / 9702 + 9899 / 9900 is calculated by recurrence equation

The above formula can be regarded as
＝1/(1×2)+5/(2×3)+11/(3×4)+19/(4×5)+29/(5×6)+…… +9899/(99×100)
……
……
＝1/(1×2)+(4+1)/(2×3)+(9+2)/(3×4)+(16+3)/(4×5)+(25+4)/(5×6)+…… +(9801+98)/(99×100)
……
……
＝(1^2+0)/(1×2)+(2^2+1)/(2×3)+(3^2+2)/(3×4)+(4^2+3)/(4×5)+(5^2+4)/(5×6)+…… +(99^2+98)/(99×100)
……
……
Each term is [n ^ 2 + (n-1)] / [n × (n + 1)]
So let an = [n ^ 2 + (n-1)] / [n × (n + 1)], and the sum of the first n terms of an is SN,
The above formula is the sum of the first 99 terms of an, that is, s99
And an = [n ^ 2 + (n-1)] / [n × (n + 1)] = [n (n + 1) - 1] / [n × (n + 1)] = - 1 / [n × (n + 1)] + 1
Let BN = - 1 / [n × (n + 1)] with the sum of the first n terms be PN and CN = 1 with the sum of the first n terms be QN, then an = BN + CN and Sn = PN + QN
QN = n; PN = - N / (n + 1)
So Sn = PN + QN = - N / (n + 1) + n = (n ^ 2) / (n + 1)
So s99 = (99 ^ 2) / (99 + 1) = 9801 / 100
The solution of PN is: BN = - 1 / [n × (n + 1)] = - [(1 / N) - 1 / (n + 1)]
So PN = - {1 / (1 × 2) + 1 / (2 × 3) + 1 / (3 × 4) + +1/[n(n+1)]}＝[1-1/2+1/2-1/3+1/3+…… +1 / n-1 / (n + 1)] = - [1-1 / (n + 1)] (where from - 1 / 2 + 1 / 2-1 / 3 + 1 / 3 +...) +1 / N can be mutually reduced) = (n ^ 2) / (n + 1)}
Note: for reference only!

2\1+6\5+12\11+20\19+30\29...+9900\9899

2/1+6/5+12/11+20/19+30/29····9702/9701+9900/9899=99-(2/1+6/1+12/1+20/1+30/1+…… +9702/1+9900/1）=99-（1-2/1+2/1-3/1+3/1-4/1+4/1-5/1+5/1-6/1+…… +98 / 1-99 / 1 + 99 / 1-100 / 1) = 99 - (1-100 / 1) = 98 and 100 / 1 = 98

2 / 1 + 6 / 5 + 12 / 11 + 20 / 19 + 30 / 29 +. + 9702 / 9701 + 9900 / 9899 Baidu didn't have any. I had to ask my own questions. I didn't get any points,
It's reversed. It should be 1 / 2, 5 / 6

The original formula is 99-1 / (1 × 2) - 1 / (2 × 3) - 1 / (3 × 4)... - 1 / (99 × 100) = 99 - [(1-1 / 2) + (1 / 2-1 / 3) + (1 / 3-1 / 4)... + (1 / 98-1 / 99) + (1 / 99-1 / 100)] = 99 - (1-1 / 100) = 98.01

It is known that the quadratic function y = ax & # 178; + BX + C (a ≠ 0) has a maximum value of 25 when x = 1 / 2, and the sum of two cubes of the equation AX & # 178; + BX + C = 0 is equal to 19

When x = 1 / 2, the function has a maximum of 25 A

Let f (x) = x2 + ax + B. for any real number x, there exists y such that f (y) = f (x) + y, then the maximum value of a is___ ．

If f (x) = x2 + ax + B, f (y) = Y2 + ay + B, then the original formula can be changed to any real number x, there exists y such that x2 + AX = Y2 + ay-y is constant, let g (x) = x2 + ax, H (Y) = Y2 + ay-y, then the range of function g (x) = x2 + ax is a subset of the range of function H (y) = Y2 + ay-y.g (x) = (x + A2)

If the quadratic function f (x) = x ^ 2 + ax + B, f (1 + x) = f (1-x) holds for any real number X
(1) Finding the value of real number a
(2) Proof: the function f (x) is an increasing function in the interval [1, + ∞)

f(1+x)=(1+x)^2+a(1+x)+b
f(1-x)=(1-x)^2+a(1-x)+b
So (1 + x) ^ 2 + a (1 + x) + B = (1-x) ^ 2 + a (1-x) + B
1+2x+x^2+a+ax+b=1-2x+x^2+a-ax+b
(4+2a)x=0
equation holds good under all circumstances
So 4 + 2A = 0
a=-2
f(x)=x^2-2x+b
Let m > n > = 1
Then f (m) - f (n) = m ^ 2-2m + B-N ^ 2 + 2n-b
=(m^2-n^2)-2(m-n)
=(m+n)(m-n)-2(m-n)
=(m-n)(m+n-2)
m>1,n>=1
So m + n > 2, M + n-2 > 0
m>n,m-n>0
So (m-n) (M + n-2) > 0
f(m)-f(n)>0
That is, when m > n > = 1
f(m)>f(n)
So f (x) is an increasing function in the interval [1, positive infinity]

Let f (x) = x ^ 2-x + A, if f (- M) < 0, then f (M + 1) is positive or negative?

f(x)=x^2-x+a,
f(-m)=m^2+m+a

Given the quadratic function f (x) = x2 + X + a (a > 0), if f (m) < 0, then the value of F (M + 1) is ()
A. The sign of negative B. positive C. 0d. Is related to a

When the function y = x2 + X is below the x-axis, - 1 < x < 0, the total interval has only one span, and ∵ a > 0 ∵ f (x) image is translated upward from the function y = x2 + X image, so the interval length less than zero will be less than 1, and ∵ f (m) < 0 ∵ m + 1 must cross the interval less than zero, so f (M + 1) must be a positive number, so choose B

Given quadratic function, f (x) = x ^ 2 + X + a (a > 0), f (m)

If f (m) = m ^ 2 + m + A0, then m ^ 2 + M

Let f (x) = x * x + X + a (a > 0), if f (m)

Choose a
F (m) = m * m + m + a < 0 and a > 0
So - M * (M + 1) > a > 0
-1< m