英文數學 求解

(1)

Suppose x is real,
x^2 + 10x + 26 = 0
(x)^2 + 2(x)(5) + (5)^2 - (5)^2 + 26 = 0
(x + 5)^2 + 1 = 0
Since the minimum of (x + 5)^2 is 0 where x is real
and the minimum of (x + 5)^2 + 1 is 1
thus, (x + 5)^2 + 1 cannot be equal to 0,
contradiction exists. so x cannot be real.

(2)

ak+1 = 3ak - 1
a1+1 = 3a1 - 1 => a2 = 3(1) - 1 = 2
a2+1 = 3a2 - 1 => a3 = 3(2) - 1 = 5
a3+1 = 3a3 - 1 => a4 = 3(5) - 1 = 14

Let P(n) be the proposition an = [3^(n - 1) + 1] / 2
when n = 1, LHS = a1 = 1 and RHS = [3^(1 - 1) + 1] / 2 = 1
thus, P(1) is true,
assume that P(k) is true for some positive integers k,
i.e. ak = [3^(k - 1) + 1] / 2
when n = k + 1,
LHS = ak+1 = 3ak - 1 = 3[3^(k - 1) + 1] / 2 - 1
= {3[3^(k - 1) + 1] - 2}/ 2
= (3^k + 3 - 2)/ 2
= (3^k + 1)/ 2
= {3^[(k + 1) - 1] + 1} / 2 = RHS, thus, P(k + 1) is true
Therefore, by the principle of mathematical induction, an = [3^(n - 1) + 1] / 2
for all positive integers n

1.
Suppose there is a real solution to the equation, call it a.

Then, a² + 10a + 26 = 0

a² + 10a + 25 = -1

(a + 5)² = -1

Contradiction!!! Because no real number can have (plus 5) then square and become negative.

Therefore, there must be NO real solution to the equation.