Answer:
a) -7/9
b) 16 / (n² + 15n + 56)
c) 1
Step-by-step explanation:
When n = 1, there is only one term in the series, so a₁ = s₁.
a₁ = (1 − 8) / (1 + 8)
a₁ = -7/9
The sum of the first n terms is equal to the sum of the first n−1 terms plus the nth term.
sₙ = sₙ₋₁ + aₙ
(n − 8) / (n + 8) = (n − 1 − 8) / (n − 1 + 8) + aₙ
(n − 8) / (n + 8) = (n − 9) / (n + 7) + aₙ
aₙ = (n − 8) / (n + 8) − (n − 9) / (n + 7)
If you wish, you can simplify by finding the common denominator.
aₙ = [(n − 8) (n + 7) − (n − 9) (n + 8)] / [(n + 8) (n + 7)]
aₙ = [n² − n − 56 − (n² − n − 72)] / (n² + 15n + 56)
aₙ = 16 / (n² + 15n + 56)
The infinite sum is:
∑₁°° aₙ = lim(n→∞) sₙ
∑₁°° aₙ = lim(n→∞) (n − 8) / (n + 8)
∑₁°° aₙ = 1
Answer:
8
Step-by-step explanation:
Let's x represents the first odd integer
The next consecutive integer would be represented as x+2.
So x and x+2 being multiplied together will give us 1443:
(x)(x+2)=1443
x^2+2x=1443
x^2+2x-1443=0
By solving the quadratic equation using the quadratic formula, x will equal to 37 and -39.
Since the problem says "integers", so I'm assuming two pairs of consecutive integers would be ok.
With that said you 1st pair will be: 37,39
Your second pair will be: -37,-39.
Answer:
-5(1-I) is the answer
Step-by-step explanation:
(1+2i)(1+3i)
1(1+3i)+2i(1+3i)
1+3i+2i+6i^2
As i^2=-1
1+5i+6(-1)
1-6+5i
-5+5i
Taking -5 as common
-5(1-i)
I hope this will help you :)
Send it to me and I'd be happy to help! :)
