FURTHER PURE MATHEMATICS (FP1) PROOF BY INDUCTION- SUMMING SERIES
PROOF OF SUMS OF SERIES BY INDUCTION
A process called mathematical induction can prove many theorems and formulae relating to
positive whole numbers. Induction can be used to prove Sums of certain series have a
particular formula although INDUCTION IS NOT ONLY USED TO PROVE THE SUM OF A
SERIES (as we will see later)
Firstly we will look at INDUCTION in a non mathematical context.
An inventor builds a climbing 'robot' which is designed to be able to climb any ladder which
has equally spaced rungs, no matter how long it may be. (Being solar-powered, it can continue
indefinitely if necessary.) The inventor tests it on a
variety of ladders: on some of them, the robot succeeds; on others it does not.
Given any ladder, what conditions need to be satisfied for the robot to be able to climb the
ladder?
You will see that there are, in fact, just two:
(a) In the first instance, the robot needs to be able to get on the ladder (presumably at the
first rung); and
(b) if the robot is on any given rung (say rung k – as rungs are equally spaced, it does not
matter where rung k is on the
ladder), then it has to be able to get on to the next rung after this (rung k +1).
If the robot's programming and construction enable it to satisfy both these conditions, then
it can climb as far up the ladder (adinfinitum if necessary) as its inventor could wish for;
since by condition (a) the robot can get on the ladder (at rung 1). Then, by condition (b) it
can get on to rung 2. By (b) again, it can get to rung 3, hence to rung 4, hence to rung 5, etc.,
etc. Thus the robot can reach rung n for any positive integer n.
WHAT HAS THIS GOT TO DO WITH PURE MATHEMATICS?!! (Good Question!)
The above is actually an illustration of a very powerful technique called proof by induction.
The method is a cunning means of proving the truth of some statement, or formula, that is
found by experimental means (for instance) but which, without a general proof, is only known
to be true for certain values of the variable concerned.
FURTHER PURE MATHEMATICS (FP1) PROOF BY INDUCTION- SUMMING SERIES
For instance, the result
This is the Sum to n terms of the natural numbers and could be proven by using
ARITHMETIC SERIES (or even the difference method) but could be proved by Induction
Consider this familiar sum from the formula booklet
however, is not an Arithmetic Series and so other methods of proof need to be investigated.
PROOF BY INDUCTION CAN BE USED TO PROVE THE SUM OF THIS (OR ANY) SERIES
FIRST PROVE TRUE FOR n=1
So the Right Hand Side (RHS) is Equal to the Left Hand Side (LHS) when n=1
(The robot is on the first rung of the ladder!)
SECONDLY MAKE THE ASSUMPTION THAT IT IS GENERALLY TRUE FOR Some Value
of n=k
This is Sometimes called THE INDUCTION HYPOTHESIS
FURTHER PURE MATHEMATICS (FP1) PROOF BY INDUCTION- SUMMING SERIES
THIRDLY USE THE ASSUMPTION TO TRY TO PROVE THAT IT IS ALSO TRUE FOR
n=K+1
There is one EXTRA term on the end of the
sum if we are now summing to K+1 terms
In this line we have USED the ASSUMPTION
from STEP 2(The induction Hypothesis)
Now look at what we were trying to prove but replace the letter n with k+1
So if the formula assumed when n = k is true then, by adding on the (k +1)th
term, it must also be true when n = k +1, whatever k is.
(The robot can get from rung k to rung k 1.)
By INDUCTION,since it was true for n=1, and it if shown true for the next integer, then it
is true for ALL POSITIVE integers
FURTHER PURE MATHEMATICS (FP1) PROOF BY INDUCTION- SUMMING SERIES
The method can be described mathematically like this.
A theorem thought to be true for all values of the positive integer n, can be proved by showing that

STEP 1 Show that it is true for some small value of n such as n = 1.(need not be 1)


STEP 2 if it is ASSUMED true for n = k,
STEP 3 and it can be shown that is also true for n= k + 1
If you can show the points above, then you have shown that the theorem is true at the start
(n = 1)
and it is therefore true of the next n value, when n = 1 + 1,
and then n = 2 + 1, and then n = 3 + 1, etc…
and so on for all integer values of n following after the valid starting value (usually n =1, but
not always).
This way of proving the validity of a theorem or formula is the method of mathematical
induction.
An essential requirement when trying to prove a proposition by induction is that you either
know the final result or can make an inspired guess and then confirm your guess by using a
proof employing induction.
EXAMPLE 2
Show by mathematical induction that
1  2.2  3.22  ......  n2n1  (n 1)2n  1
for all positive integer values of n.
SOLUTION
FIRST
Consider the theorem for n=1.
(LHS) LEFT HAND SIDE WHEN n=1
1  211
(RHS)
Which is clearly 1
RIGHT HAND SIDE WHEN n=1
(1  1)21  1 Which is again clearly 1.
So When n=1 LHS=RHS and so the theorem is true when n=1.
FURTHER PURE MATHEMATICS (FP1) PROOF BY INDUCTION- SUMMING SERIES
SECOND Assume that the theorem is true for n=k where k is some positive integer
value.
ie it is assumed true that
1  2.2  3.22  ......  n2k 1  (k  1)2k  1
This is the Induction Hypothesis and MUST be stated!!!!!!!
THIRD Consider n=k+1 (ie the next positive integer after k) and USE the ASSUMPTION
(LHS) LEFT HAND SIDE IS:
1  2.2  3.22  ......  k 2k 1  (k  1)2k
 (k  1)2  1  (k  1).2
k
k
(k  1)2k
is the next term to
be added.
The assumption made was that the theorem
was TRUE for n=k. This is the RIGHT HAND
SIDE of the theorem that is assumed to be
equal to the left hand side when n=k.
We notice that there is a common factor of
2 k between two of the terms
LEFT HAND SIDE BECOMES
 2k (k  1  k  1)  1
 2k (2k )  1
 2k 2(k )  1
 2k 1 ( k )  1
 k 2k 1  1
(RHS) NOW LOOK AT THE RIGHT HAND SIDE WHEN n=k+1
 (k  1  1)2k 1  1
 k 2k 1  1
SO When n=k+1 LHS=RHS and the theorem is true.
PUTTING IT TOGETHER
Theorem true for n=1
If true when n=k (any positive integer) then it is also true when n=k+1 (the next!)
True for n=1, so true for the next when n=2, so true for the next when n=3, so true for the next
when n=4, so, by INDUCTION is true for all positive integers
FURTHER PURE MATHEMATICS (FP1) PROOF BY INDUCTION- SUMMING SERIES
EXAMPLE 3
STEP 1 Show true for n=1
So it is true for n=1
STEP 2 Assume it is true for some integer value of n=k
STEP 3 Use the Assumption to show that if the Assumption is true then it is also true for n=k+1
Add the k+1th term
Use the Assumption
As this is the required result with n=k+1
Since it is true for n=1 and the assumption that it is true for n=k implies that it
is also true for n=k+1, then it is true for all positive integers by INDUCTION