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Summary proof by induction further maths
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Proof by induction is a mathematical technique where you prove a base case is true, then show that if the statement holds for an arbitrary ( n ), it also holds for ( n+1 ), thereby proving it for all ( n ).

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  • September 15, 2024
  • 10
  • 2023/2024
  • Summary

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k+
( + 1)(k)

.
1 (r 1)
2 r(r +
-
-
= -




In-Matthew r=1 r = 1 5

=

n(k + 1)(x 1) -
+ (k + 1)k
when n = S :
F2) +



=
k+ 1((k(x -
1) +
x) =
Exercise SA


= (2 +
1)(k)( + (x -
1) +
) 1(3(1) + 5)

② induction for
that
positive integers
RHS :I
Prove by any n
,




= In =
(k +
1)(x)( +
k -




5 +
1)
2)
~
(k + 1)(k)
(k
+
+
=



AS LHS RIS n /


BASIS
=
So =




2) (k) statement is true for
=
(k + 1) (k + >
-




n = 1 and if nik is k
Assuming
=
n

true then n = v+1 is
x(3k + 5)
1)
[(172 (1 a
soby inducti
the
Then
+
RHS =

1) (2n + 2)(2n)
.
b + (2n +
+=
(m + (m + 2)
I
=




1)(n + 2)
: true for n =
=
=(2n +

show box :

1) S)
2)
(n + 1)(3(k + +

n(n + 1)
(3r -1) =
=


2 .
ASSUMPTION : ⑭ ( +u ( + 3)
n= k
1 LHS
r(3-1)
when n
:
=


13 1)
Then
u= 1
=

+ (n +
for n = n +1


n+ 1




·
=
2
2
RHS : (1) (1 + 1)
:
show box
= 2
4


3
↑

(kT(2
=

RIS 3)
AS LHS =
So n =
1 (k + 1)(k + 2) +



=
((k2(k + 1) + ( + 2)2 =
u(3k + 5)(k + 3) + 4(k +z)

(k + 1)(k + 2)(k + 3)
Assuming
: 1
n =



(k + 1)(3k + 8)



~
. INDUCTIVE
3 then
& r(3r -

1) = n =(n + 1) =


(k + 2)(k + 3)
r = 1
x2(k + 1)
+
=
(k + 1)(3(k + 1) + 5)
=



((k + 1) +
1)((k + 1) + 2)
12 + (2 + 13
i k2(k +
=




show box : so if n = k is true then n = k + 1 is true

= = n+ 1
=
(k + 1) (in + x+ 1) (k+ 1)[(k + 2)
[r(3r 1)
=
-




(k + 1) +(x2 px 4)


so
= + + r =
1 - The statement is true for n = 1 & if n = k


the then htt is the induction
is ht
=
( +
1)2](k 2)2 +
it is true FnE I
.


re= (k + 1) (k + 2)
not
so if n = k is true for n = k + 6. .
shown for all values of htt

then v+ 1 is true is t
. re

3n2(n
=
n n+1
.




& r(zu -1)
= + 1) + (k + 1)(3(x + 1) -
1)
r = 1 v = 1




E(x + 1) + (x + 1)(3(k + 1) -
1)



~
=


3)



+
.
4 CONCLUSION b . When n = 3 (HS = (1 + = 16


The statement is true for n = 1 RHS i
2)
=
=

and if n = k is true then
=
(n + 1)(k + 3k +
,
: LHS # RIS .




n = k + 1 is the so

UnEIN = (n +
1)(k + 1)(k + 2)




~
it is true
by induction

<
(k + 1) (k + 2)



for
=
Ea ·
did not show that S = RES

1
when n =




=

m
⑤ Prove by induction
that for
any positive integer ni : the statement is true for n = 1 and if b .
CHS =




-= (n- =kiswe tennisa
RIS =
(+
LHS
r
= (n 3
Er
=




1 -
:
when n =
b
.

~
RHS -
E r(3r 1) -
= n (n + 1) A
LHS RHS
5(1)(2)(0) 0
=

r = 1




( = (+)

AS CHS = RHS n = 1 is true


1)2
+n + = 1


Assuming n = k



n2 + 2n + 1 = 4
↳ w(r -1) =

tr(u + 1)(k -
1)




~
2
n + 2n -
3 = 0
~


(n -
1)(n + 3) = 0



n = 1
or
Show box : = 3
n -




*r(r-1) =
((k + 1)(k + 2)(4) n must be +ve . :n =

r = 1

, # Matthew Exercise SB



b . 32-1 is divisible
by 8




Proving divisibility results asis
& n = 1 z -1 =
S
↑
divisible S
by
11"-6 is divisible S OneN
. Prove
eg by


Basis n =
Assume let f(n) = 3-
11 ' z2
5
-
6 =

... f(k) = -

1

divisible by S ↑
divisible by S

1)
z2(k
+

Assume f(n) 11 " 6 f(k + 1) divisible f
let Show I is
by
=
- -
=
-



for 1)
true
y2(k
+
f
(32 1)
·


n= k
Induct f(k + 1) -
f(k) = -
1 - -




so +(r) =

11-6 T
z2k 22k
+
=


divisible
by
5

f(r + D 11
**
6 is divisible
=

(32)(9) -
32
by s
=
show -




Inductive
=

32(9 -

1)
-
Step
11 k
+
(114 6)
=
32(s)
f(k + 1) -

f(x) =
-
6 - -




↑
divisible S
=
11k
+1
-
,k
by


=
(114) (11) - 112
f(k + 1) = f(k) + 324(8)
M
I
assumed divisible divisible by
=
114(11 -

1)
S S
by


~
=
(10)
↑ f(k + 1) is divisible f
by
:


divisibly by 5




1144
f(x + 1) = f(x) +
since true for n = 1 and if n = K is true then n = k+1 is true
.
,




~
↑
So induction it is true fre
assuming visible by s by
divisible
by 5

. f(k + 1) is divisible
by 5 c .
5" + q + 2 is divisible by 4




Since true for n = 1 and if n = k is true then n= k+ 1 is true
Bis n =1 5 + a + 2

so by induction
One N
=
eg.
Prove n3-In + 9 is divisible 3 OneN divisible
by 4
by

13
B n = 1
-


=(1) + 9
Assume let f(n) = 54 + qn + 2
= 1 -
7 + 9


f(k) 52 9" +
- divisible
=



is
ie .
= + 2
by 4

divisible 3
by
+ ** 1
g g
Asume f(r + 1) + 2 is divisible
+ 4
show =
by
let f(n) =
n3 -
+n + 9

ve +
·




so f(k) = 13 -
7k + 9 is divisible by 3
Induct f(k + 1) f(x) =
54
+1
+ qk + z (gk+ qk+ 2)
p
- -




=
(g4)(5) + (94)(9) + 2 54 q2 2
f(k + 1) (n + 1) = 7 (k + 1)
- -
-




+ 9 is divisible
show
by 3
=




(gn)(94)(5 2)
=

iv a +
+ 2 -

1 -
1 -




Induct + (r + ) f(r) =
(k + 1)3 =(k + 1) + 9 (x3 7u + 9)
S
- -
-




1)3 x3 +
(g2)(94) (12)
=
(x + -
7k -
7 + 9 -
7k -
9 =




=
(k + 1)3 - x3 -
7
atvisible by 4
= k3 + 342 + 34 + 1 - k3 -
7


=
322 + 3k -
6

=
3(k2 + k -


2) f(k + 1) =
f (k) + (g4)(9k)(12)
↑ ↑ ↑
assumed
divisible by 3 div
dir by 4
by 4




f(k + 1) =
f(k)
↑
+ 3(ki +
↑
x -
2) : f(k + 1) is divisible by 4 ~
assumed divisible divisible by
for 1 and if
by 3
3 since true n =
n=k is true then



~
k + 1
n =
is the OneN

: f(u + 1) is divisible
by 3

since true for n=1 B if n= K is true the n = RH is

- So
true by induction it is free for One N

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