Need Help on Geometric Sequence (Summation)

[hi25125]

Geometric Sequence (Summation)
I am told to solve the upper limit.

The lower limit is n=1
the general term is 1/2 times 3^n
the sum is 3587226
The questions says I don’t need logs to solve this. Trial and error with exponents will work.

I used this equation and tried trial and error
Tn =ar^n-1

I have tried 14.365 into the upper limit and that matches with the sum
but I think n represents the number of terms, therefore it can not be decimals.
Did I do something wrong ? Is there another way to solve this ?
Please help.

 

[galactus]

There is no decimal.

\(\displaystyle \frac{1}{2}\sum_{k=1}^{14}3^{n}=3587226\)

 

[hi25125]

how did you calculate it to be 14?
can you show me the way you put it on a scientific calculator ?

did you do the way I put it on the calculator ?
firstable, 3^14
and then divide by 2 ?
but that doesn't equal to the sum

Please help further,thank you

 

[galactus]

Just add up the powers of 3 from 1 to 14. Plug them in your calculator.

\(\displaystyle \frac{1}{2}\left(3^{1}+3^{2}+3^{3}+\cdot\cdot\cdot +3^{14}\right)=3587226\)

 

[Deleted member 4993]

Geometric Sequence (Summation)
I am told to solve the upper limit.

The lower limit is n=1
the general term is 1/2 times 3^n
the sum is 3587226
The questions says I don’t need logs to solve this. Trial and error with exponents will work.

I used this equation and tried trial and error
Tn =ar^n-1

I have tried 14.365 into the upper limit and that matches with the sum
but I think n represents the number of terms, therefore it can not be decimals.
Did I do something wrong ? Is there another way to solve this ?
Please help.

geometric series of ab + ab[sup:2tshknhs]2[/sup:2tshknhs] + ab[sup:2tshknhs]3[/sup:2tshknhs]....+ ab[sup:2tshknhs]n[/sup:2tshknhs]

\(\displaystyle S_n \ = ab\frac{b^n \ - \ 1}{b \ - \ 1}\)

3587226 = 3/2* (3^n - 1)/2

3^n - 1 = 3587226 * 4/3 = 4782968

3^n = 4782969

Now you can either keep on dividing right-hand-side by 3 - and see after how many divisions it becomes 1.

or in your calculator, keep multiplying by 3 and figure out after how many multiplications you get the number on the right-hand-side.

This is why log is used - faster and painless.....

 

[soroban]

Hello, hi25125!@

\(\displaystyle \text{Geometric Sequence (Summation)}\)

\(\displaystyle \text{The general term is: }a_n \,=\,\tfrac{1}{2}\!\cdot\!3^n\)

\(\displaystyle \text{The sum of the first }n\text{ terms is: }\,3,\!587,\!226\)

\(\displaystyle \text{Find }n.\)

\(\displaystyle \text{Formula: }\:S_n \:=\:a\frac{r^n-1}{r-1}\)

\(\displaystyle \text{We have: }\:a = \tfrac{3}{2},\;r \,=\,3\)

\(\displaystyle \text{Hence: }\:\tfrac{3}{2}\cdot\frac{3^n - 1}{3-1} \:=\:3,\!587,\!226 \quad\Rightarrow\quad \tfrac{3}{4}\,(3^n-1) \:=\:3,\!587,\!226\)

. . \(\displaystyle 3^n - 1 \:=\:4,\!782,\!968 \quad\Rightarrow\quad 3^n \:=\:4,\!782,\!969\)


\(\displaystyle \text{Divide by }3^4 = 81\!:\;\;3^{n-4} \:=\:59,\!049\)

\(\displaystyle \text{Divide by }3^4 = 81\!:\;\;3^{n-8} \:=\:729\)

\(\displaystyle \text{Divide by }3^4 = 81\!:\;\;3^{n-12} \:=\:9\)


\(\displaystyle \text{We have: }\:3^{n-12} \:=\:3^2 \quad\Rightarrow\quad n-12 \:=\:2\)

. . \(\displaystyle \text{Therefore: }\:n \:=\:14\)

 

[halcyon]

summation

Hello can anyone help me...

The question ask me to calculate the summation ∑ 4,from k=1 to 25 and ...I tried using formula when I recognize that ∑ 4 can write in ∑ 2^2..
so this is the formula I have done ...n(n+2)(2n+1)/6...but i'm not too sure about this...please someone help me...is there any easier way without using the formula?