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Series and Induction

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Introduction

#1 Series and Induction

        In this portfolio, I will be investigating a pattern and forming conjectures to explain what happens when k = 2, 3, or 4 in the 1k + 2k k + 3 k + 4 k + … + n k series. To do this, I will be using the knowledge that 1 + 2 + 3 + … + n =image00.png.

1. image01.png where a1 = 1 x 2        
                                     a
1 = 1 x 2

a2 = 2 x 3

a3 = 3 x 4

a4 = 4 x 5

.

.

.

Expression seems to show that an = n(n+1)

2. Consider Sn = a1 + a2 + a3 + … + an where ak is defined in #1 [ak = n(n+1)]

  a) Determine several values of Sk, including S1, S2, S3, …, S6

S1 = a1 = 1 x 2 = 2        

S2 = a1 + a2 = S1 + a2 = 2 + 2 x 3 = 8

S3 = a1 + a2 + a3 = S2 + a3 = 8 + 3 x 4 = 20

S4 = a1 + a2 + a3 + a4 = S3 + a4 = 20 + 4 x 5 = 40

S5 = = a1 + a2 + a3 + a4 + a5 = S4 + a5 = 40 + 5 x 6 = 70

S6 = a1 + a2 + a3 + a4 + a5 + a6 = S5 + a6 = 70 + 6 x 7 = 112

It seems that for every increasing value of k, the sum of the previous numbers plus the new number yields the new sum. All sums are multiples of 2.

b) Thus, the conjecture is that: Sn = Sn-1 + an

c) Prove conjecture by induction:

Step 1 Assume the conjecture to be true for n = 1

As shown above,

S1 = a1 = 1 x 2 = 2        

S2 = a1 + a2 = 2 + 2 x 3 = 8

S3 = a1 + a2 + a3 = 8 + 3 x 4 = 20

Step 2 Assume the conjecture to be true for n = k (done in part a of 2)        

So Sk = a1 + a2 + a3 + … + ak = Sk-1 + ak

Step 3 Observe for if n = k + 1

Should be: Sk+1 = Sk + ak+1

So substitute Sk-1 + ak for Sk (in step 2)

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Middle

1 + a2 + a3 + a4 + a5 = T4 + a5 = 210 + 5 x 6 x 7 = 420

T6 = a1 + a2 + a3 + a4 + a5 + a6 = T5 + a6 = 420 + 6 x 7 x 8 = 756

T7 = a1 + a2 + a3 + a4 + a5 + a6 + a7= T6 + a7 = 756 + 7 x 8 x 9 = 1260

It seems that for every increasing value of k, the sum of the previous numbers plus the new number yields the new sum.

b) Thus, the conjecture is that: Tk = Tk-1 + ak

c) Prove conjecture by induction:

Step 1 Assume the conjecture to be true for n = 1

As shown above,

T1 = a1 = 1 x 2 x 3= 6

T2 = a1 + a2 = 6 + 2 x 3 x 4 = 30

T3 = a1 + a2 + a3 = 30 + 3 x 4 x 5 = 90

So Tn = Tn-1 + an

Step 2 Assume the conjecture to be true for n = k (done in part a of 2)

   So Tk = T1 + T2 + T3 + … + Tk= Tk-1 + ak

   Tk = Tk-1 + ak

Step 3 Observe for if n = k + 1

According to conjecture it should be: Tk+1 = Tk + ak+1

So substitute Tk-1 + ak for Tk (in step 2):

Tk+1 = (Tk-1 + ak) + ak+1 = Tk-1 + ak + ak(a1)

d) Using the above result, calculate 13 + 23 + 33 + 43 + … + n3

Taking the sums of terms a1= 1x2x3, a2 = 2x3x4, a3 = 3x4x5, a4 = 4x5x6… from part a, it can be seen that, as it was with the differences of squares, the differences of consecutive terms (cubes) turns out to be a common number. Except this time, this number was 6 after 4 rounds of subtracting in this fashion:

 6    30    90   210      420      756      1260

    24    60    120     210       336      504

          36    60       90       126      168  

               24     30      36        42

6        6        6

The equation that I will set as my conjecture will be of degree 4, in the form Sn = an4

...read more.

Conclusion

2 + 2e + f

     = 32ª + 16b + 8c + 4d + 2e + f = 144

U3 = a35 + b34 + c33 + d32 + 3e + f

      = 243a + 81b + 27c + 9d + 3e + f = 504

U4 = a45 + b44 + c43 + d42 + 4e + f

      = 1024a + 256b + 64c + 16d + 4e + f = 1344

U5 = a55 + b54 + c53 + d52 + 5e + f

      = 3125a + 625b + 125c + 25d + 5e + f = 3024

U6 = a65 + b64 + c63 + d62 + 6e + f

      = 7776a + 1296b + 216c + 36d + 6e + f = 6048

These values were entered into the GDC as A image14.png and Bimage15.pngand it was seen by reduced row echelon form that a =image16.png, b = 2, c = 7, d = 10, e =image17.png, f = 0.

Therefore, Un = image16.pngn5 + 2n4 + 7n3 +10n2 + image17.pngn

Un = 1x2x3x4 + 2x3x4x5 + 3x4x5x6 + … + n(n+1)(n+2)(n+3)

     =image18.png= image19.png

     = image20.png

Thus, image21.png= Un – image23.png

We know that image11.png= image12.pngso image24.png=image25.png

Also, we knowimage26.png= image27.png so image28.png= image29.png and that

image30.png=6+12+18+…+6n can be rewritten by factoring out the 6 to make it 6(1+2+3+…+n). We    

already know 1+2+3+…+n=image00.png so, 6 times image00.png becomes simply 3n(n+1). Finally, we  

know that Un = image16.pngn5 + 2n4 + 7n3 +10n2 + image17.pngn so plug these values in and:

image21.png= Un – image23.png= image16.pngn5 + 2n4 + 7n3 +10n2 + image17.pngn – image25.png –    

image29.png– 3n(n+1)

= image31.png

= image32.png

Thus, my conjecture is that: image21.png= image32.png

Check:

n = 1; image21.png= 14 = 1

now try this with the above equation… image33.png= 1 [it fits!]

n = 2; image21.png= 14 + 24 = 17

now with the above equation… image34.png= 17 [it fits!]

n = 3; image21.png= 14 + 24 + 34 = 98

now with the above equation… image35.png= 98 [it fits!]

Let’s check if this conjecture works all other values of n using an induction proof:

Step 1 Assume the conjecture to be true for n = 1

As shown above,

image21.png= 14 = image33.png= 1

Step 2 Assume the conjecture to be true for n = k        

 So image37.png = image38.png

Step 3 Observe for if n = k + 1

image39.png= 14 + 24 + 34 + ... + k4 + (k+1)4 which, according to the formula, should equal

image40.png = image41.png

image42.png = image43.png+ (k+1)4 = image38.png+(k+1)4                                                         =image45.png

             = image46.png

image46.png is equal to the equation above, so

image21.png= image32.pngis true.

...read more.

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