Calculus II
posted by Sarah .
Find the sum of the series
(infinity above sigma) signma and n = 4
3^n2 divided by 2^2n3
please I really need help on this

1.125
Respond to this Question
Similar Questions

math
1)Find a1 in a geometric series for which Sn=300,r=3,and n=4 A)15 B)15/2 C)15 D)1/15 I chose A 2)Find the sum of the infinite geometric series. Sigma sign with infinity symbol above and n=1 below. To the right 20(1/4)n1 A)25 B)80/3 … 
math(need 2nd opinion)
1)Find a1 in a geometric series for which Sn=300,r=3,and n=4 A)15 B)15/2 C)15 D)1/15 I chose A 2)Find the sum of the infinite geometric series. Sigma sign with infinity symbol above and n=1 below. To the right 20(1/4)n1 A)25 B)80/3 … 
calculus
determine whether the series is convergent if so find sum: the sum of x=3 to infinity of (k+1)^2/((x1)(x2)) is it infinity meaning it diverges? 
Calculus
The problem with these two questions is that I cannot determine the a and r. The 3rd questionI don't know what I did wrong. Thanks for the help! Tell whether the series converges or diverges. If it converges, give its sum. infinity … 
Calculus
determine whether the series converges or diverges. I am stuck on these two problems. Any help is appreciated. infinity sigma (ln n)/n n=2 infinity sigma (5n^33n)/[(n^2)(n+2)(n^2+5)] n=1 
calculus
find the sum of the series: 1. the sum from n=1 to infinity of ((1)^n*(.2)^n)/n I simplified this to: (.2)^n/n I know this is alternating, but how do I know what the sum is? 
College Calculus (Binomial Series)
Expand f(x) = (x+x^2)/((1x)^3) as a power series and use it to find the sum of series (SUM from n=1 to infinity) (n^2)/(2^n) PLEASE HELP. 
Calculus
Identify the two series that are the same. a. Sigma (lower n = 4; upper infinity) n(3/4)^n b. Sigma (lower n = 0; upper infinity) (n+1)(3/4)^n c. Sigma (lower n = 1; upper infinity) n(3/4)^(n1) 
Calculus
Find the sum of the convergent series. Sigma (lower n = 0; upper infinity) 8*(4/5)^n 
Calculus
Determine convergence or divergence for the following series. State the tests used and justify your answers. Sum (infinity, n=1) 1/(1+e^n) Sum (infinity, n=1) (2*4*6...2n)/n! Sum (infinity, n=0) (n6)/n Sum (infinity, n=0) (n6)/n! …