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...(n-1)^2+n^2= [(n-1)(n-1)+2n(n-1)+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?

- Pre-cal -
**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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