SOLUTION: the nth term of a sequence is denoted by 3n(2n-1) find the sum of the first 5 terms

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Question 1205679: the nth term of a sequence is denoted by 3n(2n-1) find the sum of the first 5 terms
Found 3 solutions by MathLover1, ikleyn, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
the nth term of a sequence is denoted by 3n%282n-1%29
find the sum of the first 5 terms using n values from 1 to 5
1st term is 3%2A1%282%2A1-1%29=3
2nd term is 3%2A2%282%2A2-1%29=18
3rd term is 3%2A3%282%2A3-1%29=45
4th term is 3%2A4%282%2A4-1%29=84
5th term is 3%2A5%282%2A5-1%29=135
the sum is 3%2B18%2B45%2B84%2B135=285




Answer by ikleyn(52818) About Me  (Show Source):
You can put this solution on YOUR website!
.
the nth term of a sequence is highlight%28cross%28denoted%29%29 highlight%28cross%28by%29%29 3n(2n-1).
Find the sum of the first 5 terms
~~~~~~~~~~~~~~~~~~~~~~ 

n= 1:  a%5B1%5D = 3*1*(2*1-1) = 3*1 = 3

n= 2:  a%5B2%5D = 3*2*(2*2-1) = 6*3 = 18

n= 3:  a%5B3%5D = 3*3*(2*3-1) = 9*5 = 45

n= 4:  a%5B4%5D = 3*4*(2*4-1) = 12*7 = 84

n= 5:  a%5B5%5D = 3*5*(2*5-1) = 15*9 = 135


Thus you just know the first five terms.
To find their sum, add these 5 numbers.


You should get the ANSWER 285.

Solved.

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In whole, the task is as simple as a cucumber.

The words in the condition,  crossed by red,  are excessive and unnecessary.

They do not make the problem's formulation better.
In opposite,  they make it worst,  so I recommend to remove them.



Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Since the problem asks for the sum of only 5 terms, evaluating the five terms using the given formula and adding them is certainly one way to find the answer.

That method would be impractical if the problem asked for the sum of the first 20 terms, or the first 100 terms. For those problems you would need a more efficient method.

And you can get good practice using those methods on this relatively easy problem.



Then use the following formulas (useful in many applications):

sum%28n%5E2%2C1%2Ck%29=%28%28k%29%28k%2B1%29%282k%2B1%29%29%2F6
sum%28n%2C1%2Ck%29=%28%28k%29%28k%2B1%29%29%2F2

In this problem, those give us

sum%28n%5E2%2C1%2C5%29=%285%2A6%2A11%29%2F6=55
sum%28n%2C1%2C5%29=%285%2A6%29%2F2=15

Then the sum in the given problem is

6%2855%29-3%2815%29=330-45=285