In 1-100, what is the sum of all natural numbers with only three divisors? Urgent!

In 1-100, what is the sum of all natural numbers with only three divisors? Urgent!

Natural numbers with only three divisors must be the complete square of prime numbers
The perfect square of prime has only three divisors: 1, prime, the square of prime
In 1-100, such numbers are:
2 ^ 2, 3 ^ 2, 5 ^ 2, 7 ^ 2
Sum = 4 + 9 + 25 + 49 = 87

2 / 3 + 3x-1 / 3 x = 9x-1 / 3

Multiply by 3 (3x-1)
2(3x-1)+3x=1
6x-2+3x=1
9x=3
X = 1 / 3 (1 / 3)
Test, when x = 1 / 3, denominator is 0, that is increasing root
So the original equation has no solution

1 × 2 × 3 × 4 × 5 × 6 × 7 × 8. × 42 × 43 × 44 × 45 × 46 × 47 × 48 × 49 × 50, how much do you get

50!

It is known that an is an increasing sequence, and for any n ∈ n *, an = n & # 178; + λ n is constant, then the value range of real number λ is

We can use the idea of function to calculate an + 1

The sum of a and B is 142, the quotient of a divided by B is 6, and the remainder is 2

B = (142-2) / (1 + 6) = 20
A = 142-20 = 122

(1 / 65 + 1 / 17) * 13 + 4 / 17 simple operation
fast

=13/65+13/17+4/17
=1/5+1
=1.2

Parametric equation and conic curve
Now I feel that I have no idea about conic curve, and the amount of calculation is large. Please tell me how to use parametric equation to solve conic curve problem. Please give a detailed explanation,

Ellipse x = asin (T), y = bcos (T)
Hyperbola x = ASEC (T), y = btan (T)
Parabola x = 2pt, y = 2pt ^ 2

If M is less than N, then the system of inequalities x > m X

m

30 cm is equal to a few parts of a meter
fraction

3/10

In the sequence {an}, A1 = 1, A2 = 4, and an + a (subscript n + 1) = 4N + 1, find the formula of {an}

Brackets denote subscripts
a(n+1)=4n-a(n)+1
a(n)=a(n-1)-a(n-1)+1
Subtraction of two formulas
a(n+1)-a(n)=4-a(n)+a(n-1)
So a (n + 1) = 4 + a (n-1)
And a (1) = 1, a (2) = 4
So a (n) = 4n-3 (n is odd)
A (n) = 4N (n is even)