Help! induction question (1 Viewer)
- Thread starter maveric31
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For n = 1
Thus, the result is true for n = 1.
Assume the result is true for n = k, i.e.
^2)
(I)
We need to show the result is true for n = k+1, i.e.
^3 = (1 + 2 + ........ + k + (k+1) ) ^2)
(II)
Now,
)
^3 + (k+1)^3)
, by (I)
}{2} )^2 + (k+1)^3)
, sum of an AP series
^2 [k^2 + 4k + 4]}{4})
, from factorising (k+1)^2
^2 (k+2)^2}{4})
Also consider,
)
(k+1)}{2} ) ^2)
, sum of an AP series
Hence, LHS(II) = RHS(II).
So the result is true for n = k + 1 if it is also true for n = k.
Therefore, by mathematical induction, the result is true for all positive integers of n.
Thus, the result is true for n = 1.
Assume the result is true for n = k, i.e.
We need to show the result is true for n = k+1, i.e.
Now,
Also consider,
Hence, LHS(II) = RHS(II).
So the result is true for n = k + 1 if it is also true for n = k.
Therefore, by mathematical induction, the result is true for all positive integers of n.
Last edited:
NOTE: (1+...+k)=(k(k+1))/2 by sum of Arithmetic Progression.
First case for n=1 is trivial
Let n=k:
1^3+...+k^3=(1+...+k)^2
Let n=k+1:
LHS=1^3+...+k^3+(k+1)^3
=(1+...+k)^2+(k+1)^3
=k^2(k+1)^2/4+(k+1)^3
=RHS
RHS=((1+...+k)+k+1)^2
=(1+...+k)^2+2(k+1)(1+...+k)+(k+1)^2
=k^2(k+1)^2/4+(k+1)(k(k+1)+(k+1))
=k^2(k+1)^2/4+(k+1)^2.(k+1)
=k^2(k+1)^2/4+(k+1)^3
Above post is wrong as his case doesn't work for n=2 and has misinterpreted the question:
ie 1^3+2^3 =/= (1+2)^3
First case for n=1 is trivial
Let n=k:
1^3+...+k^3=(1+...+k)^2
Let n=k+1:
LHS=1^3+...+k^3+(k+1)^3
=(1+...+k)^2+(k+1)^3
=k^2(k+1)^2/4+(k+1)^3
=RHS
RHS=((1+...+k)+k+1)^2
=(1+...+k)^2+2(k+1)(1+...+k)+(k+1)^2
=k^2(k+1)^2/4+(k+1)(k(k+1)+(k+1))
=k^2(k+1)^2/4+(k+1)^2.(k+1)
=k^2(k+1)^2/4+(k+1)^3
Above post is wrong as his case doesn't work for n=2 and has misinterpreted the question:
ie 1^3+2^3 =/= (1+2)^3
Last edited:
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Should say ASSUME n = k is true.
Yeah, sorry. Usually i say: Assume < statement> (for n=k) but i didn't think it was necessary as i'm just writing out the proof for a person and not formally doing it in a test.