Keystone Problem… Keystone Problem… next Set 14 © 2007 Herbert I. Gross

You will soon be assigned problems to test whether you have internalized the material in Lesson 14 of our algebra course. The Keystone Illustration below are prototypes of the problems you'll be doing. Work out the problem on your own. Afterwards, study the detailed solutions we’ve provided. In particular, notice that several different ways are presented that could be used to solve each problem. Instruction for the Keystone Problem next © 2007 Herbert I. Gross

As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. next © 2007 Herbert I. Gross

next © 2007 Herbert I. Gross Find the sum of the first 16 consecutive integers, starting with 6. S = Keystone Problem for Lesson 14 That is, find the sum S, where… next

© 2007 Herbert I. Gross Solution for the Problem As a brief review, we showed in lesson 14 that… where S n is the sum of the first n positive integers; that is… next S n = n(n + 1) 2 S n = … + n

© 2007 Herbert I. Gross Solution for the Problem Thus, if the problem had been to compute the value of S 21 we simply would have replaced n by 21 in the formula S n = n(n + 1) 2 to obtain… next S 21 = 21(21 + 1) 2 = 21(22) 2 = 21(11)= 231 next

© 2007 Herbert I. Gross Solution for the Problem Notice that S (as defined above) is equal to S 21 – S 5. That is… next S = S 5 = S 21 = … + 20 S 21 – S 5 = … + 20 –

next © 2007 Herbert I. Gross Solution for the Problem next From our formula we see that… S 5 = 5(5 + 1) 2 = 5(6) 2 = 5(3)= 15 S n = n(n + 1) 2 Therefore… S = S 21 – S 5 = 231 – 15 = 216 next

© 2007 Herbert I. Gross The sequence 1, 1+2, , (that is, S1,S2, S3, S4,...) is not linear. It is “almost linear” in the sense that the increase as we go from S n to S n+1 is always n +1. next Note 1 What is really happening in this problem is that while the rate of change of S n with respect to n is not constant, the rate of change of the rate of change is constant.

next © 2007 Herbert I. Gross Note 1 More specifically the rate of change increases by 1 whenever n increases by 1. This is analogous to, say, an object moving with a constant acceleration of 1 foot per second per second. That means that the speed of the object is increasing at a rate of 1 foot per second every second. Thus, if the speed of the object is currently 1 foot per second, one second later its speed will be 2 feet per second; the next second later its speed will be 3 feet per second, etc.

next © 2007 Herbert I. Gross Note 1 This allows us to find the sum just by representing it in two different ways; once by adding the terms from left to right and once by adding them from right to left.

next © 2007 Herbert I. Gross S = … S = … S = … The sum of each pair is 27 and there are 16 pairs. More specifically, we may use the following sequence of steps... + next Note 1 Therefore, 2S = 16 × 27 = 432 Therefore, S = 432 ÷ 2 = 216

At first glance it might be difficult to recognize that the sum … consists of 16 terms.. It might be easier to notice this if we rewrite the sum in the form… (5 + 1) + (5 + 2) + … + (5 +15) + (5 + 16) …and in this form it is easy to see that the sum consists of 16 members. next