Find the sum to n terms of GP whose nth term is 3(2)^n-1
For a GP, term(n) = a r^(n-1)
comparing this with 3(2)^(n-1)
we have a = 3, r = 2
sum(n) = a(r^n - 1)/(r-1)
= 3(2^n) -1)/(1)
= 3(2)^n - 3
check:
the terms would be:
3, 6, 12, 24, ...
just adding them up, sum(4) = 45
using my formula, sum(4) = 3(2^4) - 3 = 48-3 = 45
looks good
Sure, I can help you with that! But first, let me put on my mathematical clown hat and get ready to solve this problem with a touch of humor.
So, we have a geometric progression (GP) with the nth term given as 3(2)^n-1. To find the sum to n terms, let's break it down step by step.
Now, since we're dealing with geometric progression, let's call the first term "a" and the common ratio "r". In this case, it's not explicitly given, but we can determine it by looking at the nth term.
From the nth term formula, we have:
3(2)^(n-1) = a * r^(n-1)
Now, let's find a. When n is 1:
3(2)^(1-1) = a * r^(1-1)
3 = a * r^0
3 = a * 1
a = 3
We've got the first term! Now let's find the common ratio, r. We can use the equation with n = 2:
3(2)^(2-1) = 3 * r^(2-1)
6 = 3r
r = 2
Now that we have a = 3 and r = 2, we can find the sum of the geometric progression up to the nth term.
The sum formula for a geometric series is given by:
S_n = a * (1 - r^n) / (1 - r)
So, plugging in our values, we have:
S_n = 3 * (1 - 2^n) / (1 - 2)
Simplifying this a bit, we get:
S_n = 3 * (1 - 2^n) / (-1)
And finally:
S_n = -3 * (2^n - 1)
Ta-da! The sum to n terms in the given geometric progression is -3 times the quantity (2^n - 1). Just don't take my clown calculations too seriously, as I always add a dash of humor to make things more enjoyable!
To find the sum of a geometric progression (GP) to n terms, we need to know the first term (a) and the common ratio (r).
In this case, the nth term of the GP is given as 3(2)^(n-1).
The formula for the nth term of a GP is given as:
an = a * r^(n-1)
Comparing this with the given nth term, we can see that a = 3 and r = 2.
Now, we can use the formula for the sum of a GP to n terms:
Sn = a * (1 - r^n) / (1 - r)
Substituting the values of a and r:
Sn = 3 * (1 - 2^n) / (1 - 2)
Simplifying further:
Sn = 3 * (1 - 2^n) / (-1)
Sn = -3 * (1 - 2^n)
Therefore, the sum of the given GP with n terms is -3 * (1 - 2^n).
To find the sum to n terms of a geometric progression (GP), we need to determine the common ratio (r) and the first term (a) of the series.
In this case, the nth term (Tn) of the GP is given as 3(2)^(n-1).
To find the common ratio (r), we divide the (n+1)th term by the nth term:
T(n+1) / Tn = [3(2)^((n+1)-1)] / [3(2)^(n-1)]
Simplifying this expression, we get:
2(2^n) / (2^n) = 2
So, the common ratio (r) is 2.
Now, to find the first term (a), we substitute n=1 into the given expression for the nth term:
T1 = 3(2)^(1-1) = 3(2^0) = 3
Therefore, the first term (a) is 3.
Now, we can use the formula for the sum to n terms of a GP:
Sn = a * (r^n - 1) / (r - 1)
Plugging in the values of a, r, and n, we get:
Sn = 3 * (2^n - 1) / (2 - 1)
Simplifying further:
Sn = 3 * (2^n - 1)
So, the sum to n terms of the given GP is 3 * (2^n - 1).