Solving equation x2-x + 1 = 2012

Solving equation x2-x + 1 = 2012

x²-x=2011
x²-x+1/4=2011+1/4
(x-1/2)²=8045/4
x-1/2=±√8045/2
x=(1-√8045)/2,x=(1+√8045)/2

Why is 50 square times 3 equal to 7500

=50 * 50 * 3 = 5 * 5 * 3 * 100 = 7500, no, press the computer! All these questions are asked

What is the square of (minus x) multiplied by (minus X's Square)

Negative x fourth power

The sum of the first n terms of sequence 1 / n

1 ^ 2 + 2 ^ 2 +. + n ^ 2 = n (n + 1) (2n + 1) / 6,
But there is no simple formula for 1 + 1 / 2 ^ 2 + 1 / 3 ^ 2 +. 1 / N ^ 2, but there is the formula 1 + 1 / 2 ^ 2 + 1 / 3 ^ 2 + 1 / 4 ^ 2 +. = π ^ 2 / 6

Series: 1,4,9, m, 25 Here m equals ()

m=16
It's the square of four

We know that {an} is an equal ratio sequence, A2 = 2, A5 = 14, then Sn = a1 + A2 + +The value range of an (n ∈ n *) is______ ．

∵ {an} is an equal ratio sequence, A2 = 2, A5 = 14, ∵ A5 = a2q3 = 2 × Q3 = 14 ∵ q = 12 ∵ A1 = 4, ∵ Sn = 4 × [1 − (12) n − 1] 1 − 12 = 8-8 × (12) n-1 = 8 - (12) n + 2 < 8 and ∵ A1 = 4 ∵ 4 ≤ Sn < 8, so the answer is [4, 8]

Let the sum of the first n terms of the sequence {an} be Sn = 2n & # 178; {BN} be an equal ratio sequence, and A1 = B1, B2 (a2-a1) = B1
(1) (2) let CN = an / BN, the sum of the first n terms of {CN} is TN

（1）
a(1)=S(1)=2
a(n)=S(n)-S(n-1)=2n²-2(n-1)²=4n-2,n≥2,n∈Z
When n = 1, a (1) also satisfies a (1) = 4 × 1-2 = 2
So the general formula of sequence {a (n)} is: a (n) = 4n-2, n ≥ 1, n ∈ Z
For the equal ratio sequence {B (n)}, B (1) = a (1) = 2, common ratio q = B (2) / b (1) = 1 / [a (2) - A (1)] = 1 / (4 × 2-2-2) = 1 / 4
So the general formula of sequence {B (n)} is: B (n) = B (1) × Q ^ (n-1) = 2 ^ (3-2n), n ≥ 1, n ∈ Z
（2）
For the sequence {C (n)}, the general formula is:
C(n)=a(n)/b(n)=(4n-2)/2^(3-2n)=(2n-1)·4^(n-1),n≥1,n∈Z
Let's find the first n terms and t (n) of the sequence {C (n)}:
C(1)=1·4^0
C(2)=3·4^1
C(3)=5·4^2
C(4)=7·4^3
……
C(n-1)=(2n-3)·4^(n-2)
C(n)=(2n-1)·4^(n-1)
T(n)=1·4^0+3·4^1+5·4^2+7·4^3+…… +(2n-1)·4^(n-1) ----------①
4T(n)=1·4^1+3·4^2+5·4^3+…… +(2n-3)·4^(n-1)+(2n-1)·4^n ----------②
It can be concluded from (2) to (1)
3T(n) = -1·4^0 -2·[4^1+4^2+4^3+…… +4^(n-1)] + (2n-1)·4^n
= (2n-1)·4^n - 1 - 2·[(4^n-1)/(4-1)-1]
= [(6n-5)·4^n + 5]/3
So: t (n) = [(6n-5) · 4 ^ n + 5] / 9

It is known that each item of the equal ratio sequence {an} is a positive number, Sn is the sum of the first n items, and a1 + A5 = 34, A2 · A4 = 64. Find the common ratio Q and Sn of {an}

Because the sequence {an} {an} is an equal number, so A2 · A4 = A1 · A5 = 64, and because a1 + A5 = 34, so A1 and A5 are the two roots of the equation x2-34x + 64 = 0. The solution is A1 = 2, A5 = 32, or A1 = 32, A1 = 32, A5 = 2, by an, when A1 = 2, A5 = 32, A1 = 32, A1 = 32, A1 = 32, A1 = 32, A1 = 32, A1 = 32, A5 = 32, A5 = 32, A1 = 32, or A1 = 32, A1 = 32, A1 = 32, or A1 = 32, A1 = 32, A1 = 32, Q4 = a5a1 = 16, q = 16, q = 16, q = 16, q = 2, q = 2, Sn = 2, Sn = 2, Sn = 2 (1 (1 (1 − (1 − (2n (1 − (2n) 2n = 2) (2n = 2) (1 (1 1-12n)

In known equal ratio sequence {an}, satisfy a1 + A2 = 3, A4 + A5 = 24, find the first n terms and Sn of sequence {an}

Let a (n) = k * a (n-1)
a1+a2=a1+ka1=(k+1)a1
a4+a5=k^3*a1+k^4a1=(k^3+k^4)a1=k^3(k+1)a1
Using the known A4 + A5 = 24; a1 + A2 = 3
(a4+a5 ) ：(a1+a2) = 8 =k^3
So k = 2
(k+1)a1=3
a1=1
s(n)=a1+a2+.,...+a(n)
=a1(1+k+k^2+k^3+.k^(n-1))
=1+k+k^2+k^3+.k^(n-1)
=2 ^ n-1

An = 1 / N, Sn

1+1/2+1/3+…… +1 / N is a harmonic series, there is no definite expression!
When n approaches infinity, its limit value is LN2