- Level: International Baccalaureate
- Subject: Maths
- Word count: 2305
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 =.
1. where a1 = 1 x 2
a1 = 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)
Middle
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
Conclusion
= 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 and Band it was seen by reduced row echelon form that a =, b = 2, c = 7, d = 10, e =, f = 0.
Therefore, Un = n5 + 2n4 + 7n3 +10n2 + n
Un = 1x2x3x4 + 2x3x4x5 + 3x4x5x6 + … + n(n+1)(n+2)(n+3)
==
=
Thus, = Un –
We know that = so =
Also, we know= so = and that
=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= so, 6 times becomes simply 3n(n+1). Finally, we
know that Un = n5 + 2n4 + 7n3 +10n2 + n so plug these values in and:
= Un – = n5 + 2n4 + 7n3 +10n2 + n – –
– 3n(n+1)
=
=
Thus, my conjecture is that: =
Check:
n = 1; = 14 = 1
now try this with the above equation… = 1 [it fits!]
n = 2; = 14 + 24 = 17
now with the above equation… = 17 [it fits!]
n = 3; = 14 + 24 + 34 = 98
now with the above equation… = 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,
= 14 = = 1
Step 2 Assume the conjecture to be true for n = k
So =
Step 3 Observe for if n = k + 1
= 14 + 24 + 34 + ... + k4 + (k+1)4 which, according to the formula, should equal
=
= + (k+1)4 = +(k+1)4 =
=
is equal to the equation above, so
= is true.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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