+700 I have an exam tomorrow so I am doing some practice problems, and this homework problem is really bumming me out because I keep getting weird answers!! :(
Could somebody check my work and help me with the last question regarding the sigma notation? thank you.
You come into school and tell your best friend a secret. You told her not to tell anyone but she couldn't keep it to herself, she told three other friends, but told them not to tell anyone. Will, each of those friends decided they couldn't keep it in and each told three other friends. This pattern keeps forming.
So we know that this is Geometric and I found the Explicit and Recursive functions for this problem (hopefully they are right)
\(f(n)=1(3)^{n-1}\)
\(f(1)=1 \)
\(f(n)= f(n-1)3\)
Then it asks how many people will be told on day 6 (should we use recursive or explicit)? and i said explicit
Give the first terms of the sequence (should we use recursive or explicit)? i said explicit
Use Sigma Notation to solve. What is the total number of people that will know the secret on day 40?
\(\sum_{n=1}^{40} 1(3)^{n-1} \)
I plugged in the explicit funtion i found and I got stuck here. I got a scientific notation number but it does not give me the exact values. Did I do this problem right? Can someone please please please help me.
Thank you so much in advance!!!
Here is my attempt:
1, 3, 9, 27, 81, 243, 729........etc.
3^0, 3^1, 3^2, 3^3, 3^4, 3^5, 3^6......etc.
On day 6, there will be:3^(6 -1) =3^5 =243 people.
Total number of people by the 6th day =S(n) =a(1) * (1 - r^n) / (1- r) =1 * (1 - 3^6) / (1 - 3) =1 * (-728 / -2)= 364 people, or:{1 + 3 + 9 + 27 + 81 + 243} =364 people.
Or: ∑[3^(n - 1), n, 1, 6] = 364
On the 40th day =3^(40 - 1) =3^39 =4,052,555,153,018,976,267 people !!!.
Your last summation: ∑[3^(n - 1), n, 1, 40] = 6078832729528464400 - This is total number of people from day 1 through 40th day.
+118609 Thanks guest
Give the first terms .. you could use either but recursive would be easier because you just have to keep multiplying by 3.
You sum is correct.
I just used the sum of a GP to solve it
\(Sum=\frac{a(r^n-1)}{r-1}\\ Sum=\frac{1(3^{40}-1)}{3-1}\\ Sum=\frac{3^{40}-1}{2}\\ Sum\approx 6.0788\times 10^{18}\)
