You can
put this solution on YOUR website!"x mod y" is short for "remainder of the division of x by y". Iff "x mod y" is 0, x is divisible by y. In our case, y = 9.
(mod 9) =
(mod 9)=
(mod 9) =
(mod 9)
9n^2 is always divisible by 9, so we shouldn't worry about that term. The same goes for the "+9" at the end, and the "+9n" at the middle.
(mod 9) =
(mod 9)
Therefore,
P(x) =
(mod 9) = 0
is equivalent to
S(x) =
(mod 9) = 0
Now suppose S(n) is true. I'll try to prove that S(n+1) is also true.
(mod 9) =
(mod 9) =
(mod 9) =
(mod 9) =
(mod 9)
But wait, if S(n) is true, then
(mod 9) =
(mod 9)
The inductive step is done. Now we need to check the base case of natural numbers (which is 0)
(mod 9) =
(mod 9)
S(x) is therefore true for all naturals.
You can
put this solution on YOUR website!We prove it through mathematical induction
for n =1
P(n)= 1^3 + 2^3 + 3^3
= 36 which is divisble by 9
hence P(1) is true
Let it be true for n=m
P(m) is true
We have to now prove that it is true for n=m+1
given P(m+1) = (m+1)^3 + (m+2)^3 + (m+3)^3
= (m+1)^3 + (m+2)^3 + m^3 + 9m^2 + 27m + 27 Expanding the last term
= [ m^3 + (m+1)^3 + (m+2)^3 ] + 9m^2 + 27m + 27 Rearranging and grouping
= P(m) + 9(m^2+ 3m +3)
the expression above is also divisible by 9 as P(m) is divisble by 9 and
9(m^2+ 3m +3) is divisble by 9.
Hence it is true for all n . Proof by induction complete
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