+1124 Hi friends,
I have stumbled upon a problem which I believe is not solveable, am I doing something wrong or is it one of those sums?
Above the Sigma notation we have "m"
Bottom we have k=1
The expression is: \((2)^{8-k}=255{1 \over2}\)
Determine "m"
Okay, I calculated T1, and got 128
T2, and got 64
Since it is stated that this is an Arithmetic sequence, I know that d = -64
Using \(Sn={n \over2}[2a+(n-1)d]\)
I get to a step where it says
\(64n^2-320n+511=0\)
This is not solveable...I'm I doing something wrong here?..Please help. Thank you.
a=listfor(k, 1, 9, (2^(8 - k))
The above is your sequence. It is NOT arithmetic sequence but a GEOMETRIC sequence, where:
F==128 - this the first term, m ==9 - this is the number of terms, r ==1/2 - this is the common ratio.
So, your GS looks like this:
(128, 64, 32, 16, 8, 4, 2, 1, 0.5)==255.5 {As you can see here, r ==1/2}
You sum it up using the sum formula for GS:
{128 * [( 1 - 1/2^9) / (1 - 1/2)]}==S
{128 * [(1 - 0.001953125) / (1/2)]}==S
{128 * [0.998046875 / 0.5]} ==S
{128 * 1.99609375} ==255.5 - which is the answer you want.
+1124 Hi guest,
Thank you...yes the paper indicated this was an Arithmetic sequence...no wonder I got stuck...Thank you.
+118673 Hi Juriemagic,
\(\displaystyle \sum _{k=1}^m\;\;2^{8-k}=255.5\)
First I did a little spreadsheet to see if it worked
You can see that the answer is 9.
Now I will try it without the help of Excel.
The sequence is 128, 64, 32, 16 ,....
This is a GP with r=-0.5 and a=128
\(S_n=\frac{a(1-r^n)}{1-r}\\ S_n=\frac{128(1-(0.5)^n)}{0.5}=255.5\\~\\ 256(1-(0.5)^n)=255.5\\~\\ 1-(0.5)^n=\frac{255.5}{256}\\~\\ 1-\frac{255.5}{256}=(0.5)^n\\~\\ \frac{0.5}{256}=\frac{1}{2^n}\\~\\ \frac{1}{512}=\frac{1}{2^n}\\~\\ 2^n=512\\~\\ n=9 \)
+1124 Hello Melody,
mmm, I know I would have gotten it right if it was'nt for the misleading info that this was an Arithmetic sequence. Let this be a lesson to ALWAYS calculate the 3rd term as well and MAKE SURE the sequence is indeed what the paper says it is. Thank you very much Melody...I do appreciate..
