Posted by Lucy on Saturday, January 3, 2009 at 11:46am.
use mathematical induction to prove that 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6 for all positive integral values of n.
My work: n=1: 1(1+1)(2(1)+1)/6 = 1(2)(3)/6 = 6/6 = 1.
1^2+2^2+3^2...(n1)^2+n^2= [(n1)(n1)+2n(n1)+1]/6 = 2n^3+3n^2+n/6 = n(2n=1)(n+1)/6 = n(n+1)(2n+1)/6. Since Sn is valid for n=1, it is also valid for n=2, n=3, etc.
The teacher marked this as wrong but this followed the example provided by the book. Where did I go wrong?

Precal  Reiny, Saturday, January 3, 2009 at 12:10pm
Step 1
show it to be true for n=1
LS = 1^2 = 1
RS = (1)(2)(3)/6 = 1, checks out
Step 2
Assume it is true for n=k
that is..
1^2 + 2^2 + 3^2 + ... + k^2 = k(k+1)(2k+1)/6
Step 3
is it then true for n = k+1 ??
or
is 1^2 + 2^2 + 3^2 + ... + k^2 + (k+1)^2 = (k+1)(k+2)(2k+3)/6 ??
from Step 2 the sum of terms to the end of k^2 = k(k+1)(2k+1)/6
then
LS = k(k+1)(2k+1)/6 + (k+1)^2
= (k(k+1)(2k+1) + 6(k+11)^2)/6
= (k+1)(k(2k+1) + 6(k+1))/6
= (k+1)(2k^2 + k + 6k + 6)/6
= (k+1)(k+2)(2k+3)/6
= RS
Q.E.D.
compare your solution to mine.
Where you not shown the 3step process of an induction proof?
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