The square of 100 pie minus r equals 200

The square of 100 pie minus r equals 200

The square of 100 pie minus r equals 200
Square of 100-r = 200 △ π
Square of 100-r = 63.69
R squared = 36.31
r≈6.03

The square of a few minus the square of a few equals 100

14.5²-10.5²=100
26²-24²=100

How to solve the problem that the square of x equals 20x + 200

X^2=20X+200；
Therefore: x ^ 2-20x-200 = 0;
According to the root formula: x = [- B ± √ (b ^ 2-4ac)] / (2a)
The results show that: X1 = 10 + 10 √ 3; x2 = 10-10 √ 3;

Sum: 1 * 4 + 2 * 7 + 3 * 10 +. + n (3N + 1)

N * (n + 1) ^ 2 do you want to simplify the above expression? If so, the answer is what I wrote above. Note that it's n times the square of (n + 1)~

Sum, 2 + 2 ^ 4 + 2 ^ 7 +... + 2 ^ 3n-1 =?

It should be 2 + 2 ^ 4 + 2 ^ 7 +... + 2 ^ (3N + 1)
This is an equal ratio sequence, and q = 2 ^ 3 = 8 is not equal to 1
So the original formula = 2 * (1-8 ^ n) / (1-8) = 2 * (8 ^ n-1) / 7

If an = (1 / 2) ^ n + 3n-2, find the first n terms and Sn of the sequence=

An = an arithmetic sequence plus an arithmetic sequence
Sn = sum of an arithmetic sequence and sum of an arithmetic sequence
=1-(1/2)^n+(3n-1)n/2
=(3n+1)n/2-(1/2)^n

Given that the sequence an is an arithmetic sequence and a1 + A2 + a3 = 2 π, then Tan (A3 + A5) =?
How can we work out the tolerance?

a1 a1 d a1 2d=2π,
3a1 3d=2π,
a1 d=2π/3.==>a2=2π/3.
a1 a1 d a1 2d=a1 d 2a2=3a2=2π.
a3 a5=a1 2d a1 4d
=2a1 6d.
=2(a1 3d)=2a4.
… Bad condition, A1 =?

If the first n terms of an, BN and Sn, TN satisfy Sn / TN = 3N + 1 / 2n + 5, then A3 / B3=

S5=5(a1+a5)/2=5*(2a3)/2=5a3
Sn/Tn=3n+1/2n+5,
a3/b3=S5/T5=(3*5+1)/(2*5+5)=16/15,

If Sn / TN = 2n / 3N + 1, then what is A3 / B3?

a3/b3
=2a3/2b3
=(a1+a5)/(b1+b5)
=[(a1+a5)*5/2]/[(b1+b5)*5/2]
=S5/T5
=2*5/(3*5+1)

If the first n terms of an, BN and Sn, TN satisfy Sn / TN = 3N + 1 / 2n + 5, then A5 + B3=
Sorry, it's A5 / B3=

The sum of arithmetic sequence can express not (first term + last term) × number of terms / 2
Sn=(a1+an)*n/2
Tn=(b1+bn)*n/2
(a1+an)/(b1+bn)=(3n+1)/(2n+5)
Let a1 + an = k * (3N + 1)
b1+bn=k*(2n+5)
an= a1 + q1(n-1)
The solution is an = k * (3n-1)
bn=k*(2n+3/2)
A 5 / B 3 were substituted into 28 / 15, respectively