An Interesting Application of Series and SequencesAuthor(s): Nancy S. RosenbergSource: The Mathematics Teacher, Vol. 76, No. 4, Gifted Students (April 1983), pp. 253-255Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27963461 .

Accessed: 18/07/2014 09:55

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org.

.

National Council of Teachers of Mathematics is collaborating with JSTOR to digitize, preserve and extendaccess to The Mathematics Teacher.

http://www.jstor.org

This content downloaded from 129.130.252.222 on Fri, 18 Jul 2014 09:55:09 AMAll use subject to JSTOR Terms and Conditions

sharing leaching ideas

An Interesting Application of Series and Sequences

Most students of advanced algebra are

acquainted with the fact that S2. The feth term of this series can be ex

pressed as

l2 + 22 + + lX2n + 1).

Frequently they are asked to prove it with

mathematical induction. The usual deri

vation of formulas like this one involves

the use of the binomial theorem. Recently, my students and I have developed an in

teresting alternate proof that uses series

and sequences. In the course of discussing arithmetic

series and their sums, we proved that the sum of the first odd numbers is n2. To

illustrate the pattern, I wrote several equa tions on the board:

12 = 1

22 = 1 + 3 32 = 1 + 3 + 5 42 = 1 + 3 + 5 + 7

Suddenly I realized that by adding equa tions like these it should be possible to

arrive at the formula for the sum of the first

squares. Letting S2 represent this sum we

wrote the configuration in table 1. We now

focused on the series on the right side of

Thus,

S2 =

s2 =

or

[n-(fc-l)][2fc-l].

E[n-(k-l)][2*-l], *=i

M2* k=l

*? - *=1

-l)-(*-lX2*-l)],

n-2fc2 + 3fc- 1],

S2 = ?[fc(2n + 3)-2fc2-(n+ 1)]. fc=l

Since

and

cak = c ? ak

fc= 1 k=l

c = nc, * = i

then S2 can be written as

S2 = (2n + 3) ? * - 2 ? *2 - + )?

TABLE 1

12= 1

22 = 1 + 3

32 = 1 + 3 + 5 42 =1 + 3 + 5 + 7

n2 = 1+ 3 + 5 + 7 + ???

+ (2n ?

1)

52 = l2 + 22 + 32 + 42 +

? ? ? + n2 = n(l) + (? -

1X3) + ( -

2X5) + ? ? ? + ?(2n

- 1)

Sharing Teaching Ideas offers practical tips on the teaching of topics related to the secondary school

curriculum. We hope to include classroom-tested approaches that offer new slants on familiar subjects for the

beginning and the experienced teacher. Please send an original and four copies of your ideas to the managing editor for review.

April 1983 253

This content downloaded from 129.130.252.222 on Fri, 18 Jul 2014 09:55:09 AMAll use subject to JSTOR Terms and Conditions

The first of these summations is equal to

(n/2Xn + 1); the second is S2, the very sum we are seeking. Substituting, we get

(1) n3=(

?(?+!)

2n + 3 2

2n + 3

- 2S2

- + 1),

- + 1),

- 1

-2"

S2 = (2n + 3)

3S2 = (2n + 3)

3S2 = n(n + 1)

3S2 = + 1)

and

S2=-(n+Wn+i).

Fresh from this victory, we undertook to derive the formula for the sum of the first cubes. Pursuing the analogy, our first task was to express n3 as the sum of a series of

integers.

The starting point for our first derivation had been the observation that

n2 = 1 + 3 + 5 + 7 + ? ? ? + (2n -

1).

This equation resulted from applying the formula for the sum of terms of an arith

metic series,

S = ( r fri + a

to the first odd numbers, that is, to the case where tx = 1 and tn = 2n ? 1. This ap proach had yielded

S = I 2 ICI + (2? 1)] = ( \ )[2n] = n2.

If the formula for the sum of the first

squares was based on the fact that

n2 = ( -2 l[l + (2n Dl

then a formula for the sum of the first cubes could be based on the fact that

[1 + (2n2 - 1)].

The right side of (1) is a formula for the sum of an arithmetic series with it = 1, and

tn = 2n2 ? 1. Using this value of tn in the formula for the nth term of an arithmetic

sequence,

tn = h + ( -

IK

we found that for > 1, d, the difference between two successive terms of the se

quence, was In + 2. This observation was used to generate the following set of equa tions:

l3

23

33

43

53

63

1

1 + 7 1+9+17 1 + 11+21 + 31 1 + 13 + 25 + 37 + 49 1 + 15 + 29 + 43 + 57 + 71

Adding the columns on the right is a much more difficult task in this case since each column is itself an arithmetic se

quence. The third column, for example, equals 17 + 21 + 25 + 29 + ? ? ?. Table 2 is an analysis of the form these columns take.

Substituting in the formula for the sum

of an arithmetic series,

S = ( 2ph + (n

- !)</],

we found the sum of the series in the kth

column to be as follows :

[^f^][2[l + 2(/c2-l)]

+ [n-(fc-l)-l][2(fe-l)]]

TABLE 2

Column Number_1_2_3_4_5_k_

First term = tt 1+0(1 + 1) 1+2(2+1) 1+4(3+1) 1 + 6(4+1) 1 + 8(5+1) 1 + 2(k - l\k + 1)

Difference = d 0 2 4 6 8 2(/c ? 1) Number of terms = ? 1 ?2 ? 3 ? 4 ?

{k ?

1)

254 Mathematics Teacher

This content downloaded from 129.130.252.222 on Fri, 18 Jul 2014 09:55:09 AMAll use subject to JSTOR Terms and Conditions

[2[2fc2 -

1]

+ [ - *][2(* - 1)]] = (

- k + l)?2k2 - l + (

- fcXfc

_ 1)]

= (?-* + 1)[2 2 - 1 + nk - k2 - + fc]

= 'k3 + k(n2 + 3? + 2) - (n2 + 2n + 1) = ~k3 + (n + IX? + 2)fc - ( + l)2.

The sum of the first cubes, S3, is found

by taking the sum of this quantity as k runs from 1 to n. Thus,

S3 = 13 + 23 + 33 + 43 + ??? +

3

= E"*3 + (? + IX? + 2)fc - ( + l)2]

= - k3 + ( + 2\ + 1) *

+ I)2

= - S3 + ( + 2 ? + 1)

- ( + 1)

-2.J

So

2S,

2S, =

S, =

= ?( + 1){^

- ( + I)2

- 1

+ I)2

2( + I)2

In theory, it should be possible to extend this method once more to obtain a formula for the sum of the fourth powers of the first

integers. In practice, however, the diffi culties are enormous. It now becomes nec

essary to add sequences of numbers whose differences themselves form arithmetic se

quences ! For purposes of comparison, the usual

derivation of the formula for the sum of the first cubes is given below. It is far more

elegant than the one presented here, but

correspondingly more contrived. From the binomial theorem, we know

that

(k + l)4 - kA = 4k3 + 6k2 + 4* + 1.

Allowing k to assume integral values from 1 to n, we have

k = 1

k = 2

24-14 = 4?13 + 6?12 + 4?1 + 1

34 - 24 = 4?23 + 6?22 + 4?2 + 1

k = n

( + l)4 - 4 = 4 ? n3 -h 6 ? 2 + 4 ? + 1

These equations are added to give

(n + \ f - l = 453 + 6 ? k2 + 4 ? k + k=l k=l

where S3 is again the sum of the first cubes. Then

n4 + 4n3 + 6n2 + An = 4S3

+ 6^

{In + lXn + 1)J + 4^

(1 + n)J

+ n.

Simplifying and solving for S3 yields the same result as before.

The formula for the sum of the first

squares can be derived in a completely analogous manner, and a generalization of this method yields the following formula for

Se_! = lfl_1 + 2fl_1 + 3a"1 +

??? + nfl-1:

: "?+(>

+ Ir'] 3

+ ...

+ a

a - 1

In theory, this formula can be used recur

sively to calculate the sum of any desired

powers.

Nancy S. Rosenberg Riverdale Country School

Bronx, NY 10471

April 1983 255

This content downloaded from 129.130.252.222 on Fri, 18 Jul 2014 09:55:09 AMAll use subject to JSTOR Terms and Conditions