# Series and Induction

#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 =.

1.  where a1 = 1 x 2
a
1 = 1 x 2

a2 = 2 x 3

a3 = 3 x 4

a4 = 4 x 5

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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) and k(k+1) for ak where n = k (refer to #1):

Sk+1 = a1 + a2 + a3 + … + ak + ak+1

= Sk-1 + ak + ak+1 = Sk + ak+1

We got the answer that we wanted, so the conjecture is true!

d) Using the above result, calculate 12 + 22 + 32 + 42 + … + n2

It can be seen that the differences of consecutive terms (squares) turns out to be a common number of 2:

12   22    32   42   52    62   72

3     5     7    9    11    13

2    2     2    2      2

Also, taking the terms a1= 1 x 2, a2 = 2 x 3, a3 = 3 x 4, a4 = 4 x 5… from #1, the differences of consecutive terms also turns out to be a common number of 2:

2   6    12    20    30   42

4    6     8    10     12

2    2     2      2

Taking the sums of these terms that were investigated in part a, it can be seen that:

2    8     20    40     70    112

6   12     20    30     42

6      8      10     12

2       2       2

In the last case, the numbers in my first two rows of differences are not all the same, but in the third row of differences, the differences are all the same. Because the difference is 2 after three rounds of subtracting in this fashion, the equation that I will set as my conjecture will be of degree 3, in the form Sn = an3 + bn2 + ...