Let denote the -th term of the sequence, starting with the index . Denote by the forward differences of , that is
for . We have
which indicates is an arithmetic sequence with common difference between terms of 3 and starting with 4, so that
for . So we have
and so on, with the th term given by
Reversing the order of all but the first term in the sum gives
so that
for .
Answer:
2 and 11
Step-by-step explanation:
Let's start by calling these two numbers x and y. Now, you can set up the following system of equations:
x*y=22
x+y=13
You can rearrange the following equation to be x=13-y, and substitute into the first.
(13-y)*y=22
-y^2+13y=22
y^2-13y+22=0
y=2 or 11
x=11 or 2
Hope this helps!
Answer:The whole numbers are the numbers 0, 1, 2, 3, 4, and so on (the natural numbers and zero). Negative numbers are not considered "whole numbers." All natural numbers are whole numbers, but not all whole numbers are natural numbers since zero is a whole number but not a natural number.
Step-by-step explanation:
Answer: x = (3 − i√3)/6 and (3 + i√3)/6
Step-by-step explanation:3x² − 3x + 1 = 0
∴ 3x² − 3x = -1
∴ x² − x = -1/3
∴ x² − x + (-1/2)² = (-1/2)² − 1/3
given x² + bx + (b/2)² = (x + b/2)²
∴ (x − 1/2)² = 1/4 − 1/3
∴ (x − 1/2)² = 3/12 − 4/12
∴ (x − 1/2)² = -1/12
∴ x − 1/2 = ±√(-1/12)
This tells us there are no real roots and if you need real number solutions we stop here
∴ x − 1/2 = ±i/(2√3)
∴ x − 1/2 = ±i√(3)/6
∴ x = 1/2 ± i√(3)/6
∴ x = 3/6 ± i√(3)/6
∴ x = (3 − i√3)/6 and (3 + i√3)/6 <= the 2 complex roots