Algebra II / Mr. Hansen
3/16/2006

Name: ___________KEY___________

Check for Understanding: Chapter 11

A calculator is required. This quiz will not be graded, but I want everyone to take it for educational and diagnostic purposes.

 

1.

Match each formula to its purpose and fill in the blanks.

 

 

 

 

t1 + (n – 1)d

= nth term of an arithmetic sequence (or series)

 

t1(1 – rn – 1)/(1 – r)

= nothing

 

t1rn – 1

= nth term of a geometric sequence (or series)

 

n((t1 + tn)/2)

= nth partial sum of an arithmetic series

 

t1/(1 – r)

= infinite sum of a geometric series where |r| < 1

 

t1(1 – rn)/(1 – r)

= nth partial sum of a geometric series

 

 

 

2.

Add up this series: 3 – 3/2 + 3/4 – 3/8 + 3/16 – 3/32 + . . .

 

S₯ = t1/(1 – r) = 3/(1.5) = 2

 

 

3.

What term number is –3/2048 in #2?

 

Solve this equation for n: t1rn – 1 = –3/2048. Or, if you prefer, count on your fingers and toes. Either way, the answer is 12. There is no credit in this case for being off by 1, since an odd number is not a good guess. (Only the even numbered terms are negative.)

 

 

4.

Compute the sum of the 43rd through 59th terms of the series in #2.

 

S59 – S42 = t1(1 – r59)/(1 – r) – t1(1 – r42)/(1 – r)
   = 3(1 – (–½)59)/1.5 – 3(1 – (–½)42)/1.5
   = 3(1 + 0.559)/1.5 – 3(1 – 0.542)/1.5
   = 2(1 + 1.7347 . . . (10–18)) – 2(1 – 2.2737 . . . (10–13))
   » 4.548 · 10–13

Note: I think that in E period, the answer I wrote on the board had an incorrect negative sign. The answer shown above is correct. We know that the sum of the 43rd through 59th terms must be positive, since t43 is positive and dominates all the terms that follow it.

Alternate method proposed by Peter H.: Create a new geometric series that starts with the old 43rd term and goes for 17 terms, through the old 59th term. Although we would probably never want to write it out by hand, this series looks like this:

3(–½)43 + 3(–½)44 + 3(–½)45 + 3(–½)46 + 3(–½)47 + 3(–½)48 + 3(–½)49
   + 3(–½)50 + 3(–½)51 + 3(–½)52 + 3(–½)53 + 3(–½)54 + 3(–½)55 + 3(–½)56
   + 3(–½)57 + 3(–½)58 + 3(–½)59

Because Peter’s series is geometric, with first term 3(–½)42 and common ratio –½, we can use our formula t1(1 – rn)/(1 – r) to find the sum of the first 17 terms, except with t1 = 3(–½)42 this time instead of 3.

Answer: t1(1 – rn)/(1 – r) = 3(–½)42(1 – (–½)17)/1.5 » 4.548 · 10–13. Bravo, Peter!

 

 

5.

Prove that the sum of a finite arithmetic series of n terms equals n/2 times the sum of the first and last terms.

 

Given: Let t1 = a Ξ Β, common difference d Ξ Β, and n = # of terms
Prove: Sn = n/2 · (t1 + tn)
Proof (in the style of the young Gauss):

Sn = (a) + (a + d) + (a + 2d) + (a + 3d) + . . . + (tn – d) + (tn)
Sn = (tn) + (tn – d) + (tn – 2d) + (tn – 3d) + . . . + (a + d) + (a)
_________________________________________________
Add equations to get the following:
2Sn = (a + tn) + (a + tn) + (a + tn) + (a + tn) + . . . + (a + tn) + (a + tn), for a total of n addends

Therefore, 2Sn = n(a + tn), from which the conclusion follows by dividing both sides by 2 and substituting t1 for a. (Q.E.D.)