n an sum n
1 2 2
2 6 8
3 10 18
4 14 32
n (4n-2) 2n^2
(n+1) (4(n+1)-2) [2n^2+ (4(n+1)-2)] or hopefully 2(n+1)^2
[2n^2+ (4(n+1)-2)] = 2n^2+4n +2
and
2(n+1)^2 = 2(n^2+2n+1) = 2n^2+4n+2 done
2n^2+ (4(n+1)-2)
Use Mathematical Induction to show that the statement
2 + 6 + 10 + . . . + (4n – 2) = 2n^2
is true
2 answers
Using the 3 step process:
1. test for n=1
LS = 2
RS = 2(1^2) = 2
2. assume it true for n=k
that is ....
2+4+6 + ... + (4k-2) = 2k^2
3 . prove it is then true for =k+1
or
2+4+6+ ... + (4k-2) + 4(k+1)-2 = 2(k+1)2
LS = [2+4+6+...+ 4k-2 ] + 4(k+1)-2
= 2k^2 + 4k + 4 - 2
= 2(k^2 + 2k + 1)
= 2(k+1)^2
= RS
QED
1. test for n=1
LS = 2
RS = 2(1^2) = 2
2. assume it true for n=k
that is ....
2+4+6 + ... + (4k-2) = 2k^2
3 . prove it is then true for =k+1
or
2+4+6+ ... + (4k-2) + 4(k+1)-2 = 2(k+1)2
LS = [2+4+6+...+ 4k-2 ] + 4(k+1)-2
= 2k^2 + 4k + 4 - 2
= 2(k^2 + 2k + 1)
= 2(k+1)^2
= RS
QED
1 answer