Answered By Nahome D

1.

Geometric Series

 $\sum_{k=0}^{n-1}ar^k=a\left(\frac{1-r^n}{1-r}\right)$∑_k=0ⁿ⁻¹ar^k=‍a(1−rⁿ1−r ) 

a=r=2 and the entire sum is equal to 126.

 $2\left(\frac{1-2^n}{1-2}\right)=126$2‍(1−2ⁿ1−2 )=126 

 $2^n-1=63$2ⁿ−1=63 

 $2^n=64$2ⁿ=64 

 $n=\frac{ln64}{ln2}$n=ln64ln2  

 $n=6$n=6 

Answered By Eric C

3. From the series:  $5+13+21+...+149$5+13+21+...+149 we can see that, starting at 5, each term is equal to the last term plus 8.

Hence,  $\sum_{i=1}^n\left[8\left(n-1\right)+5\right]$∑_i=1ⁿ[8(n−1)+5] 

For this case, you can easily find the number of terms (n) by subtracting the first term from the last term (ie. 149-5=144) then dividing by the increment (ie. 144/8=18). This is the number of terms minus the first term (n-1=18). So by adding 1, we get n=19

The sum of the series is calculated by substituting n with the calculated number of terms (ie. 19)

Hence, $\sum_{i=1}^{n=19}\left[8\left(n-1\right)+5\right]=1620$∑_i=1ⁿ⁼¹⁹[8(n−1)+5]=1620 

Answered By Eric C

Sorry, for question 3. The final answer is 1463, not 1620

Answered By Harrison V

1. For the sum of a geometric series, where 'r' is our ratio and 'a' is our first term, we have:    $\sum_{i=0}^{n-1}ar^i=a\left(\frac{1-r^n}{1-r}\right)$∑_i=0ⁿ⁻¹arⁱ=a(1−rⁿ1−r )   

Note that if we want our 1st term we plug in i=0. Perhaps that feels a little strange, but if we do so then our first term is $a\cdot r^0=a$a·r⁰=a, which is what we'd expect. Now, plugging a=r=2 and letting the whole equation to equal 126, we find:

 $a\left(\frac{1-r^n}{1-r}\right)\rightarrow2\left(\frac{1-2^n}{1-2}\right)=126$a(1−rⁿ1−r )→2(1−2ⁿ1−2 )=126 

   $\left(\frac{1-2^n}{1-2}\right)=63$(1−2ⁿ1−2 )=63 

 $2^n-1=63$2ⁿ−1=63 

 $2^n=64$2ⁿ=64 

 $log\left(2^n\right)=log\left(64\right)$log(2ⁿ)=log(64) 

  $nlog\left(2\right)=log\left(64\right)$nlog(2)=log(64) 

 $n=\frac{log\left(64\right)}{log\left(2\right)}=6$n=log(64)log(2) =6 

So this series must have 6 terms, though we started at n=0. 

2. We want to know what the 7th term is (IOW: when n=7), and r=a=2.

  $ar^i\rightarrow2\cdot\left(2\right)^6=128$arⁱ→2·(2)⁶=128 

3.   We first need to calculate the number of terms in the sequence. We could use a formula, but we can also just hit it with the logic-stick and see what happens. Clearly, this series increases by 8 each term. We can determine the number of a term by subtracting the first term (5) and dividing by the common difference (8).

 $\frac{\left(5-5\right)}{8}=0$(5−5)8 =0 

 $\frac{\left(13-5\right)}{8}=1$(13−5)8 =1 

 $\frac{\left(21-5\right)}{8}=2$(21−5)8 =2 or

 $\frac{\left(149-5\right)}{8}=18$(149−5)8 =18 

Judging by the numbering, we need to add (1) to our value because these numbers we've calculated are off by one. So, if 5 is the 1st term, then 149 is the 19th term. 

To calculate the sum of an arithmetic series, we use the following equation, (where  $a_1=5$a₁=5  and n is the term number, n=19)

 $\frac{n\left(a_1+a_n\right)}{2}\rightarrow\frac{19\left(5+149\right)}{2}=1463$n(a₁+a_n)2 →19(5+149)2 =1463 

Therefore, the sum of our arithmetic series is 1463.