Question

2. For every value of n, 1n
=​

Answer 1

Step-by-step explanation:

Annotated Example of Mathematical Induction

Prove 1 + 4 + 9 + ... + n2 = n (n + 1) (2n + 1) / 6 for all positive integers n.

Another way to write "for every positive integer n" is for every positive integer n. This works because Z is the set of integers, so Z+ is the set of positive integers. The upside down A is the symbol for "for all" or "for every" or "for each" and the symbol that looks like a weird e is the "element of" symbol. So technically, the statement is saying "for every n that is an element of the positive integers", but it's easier to say "for every positive integer n".

Identify the general term and nth partial sum before beginning the problem

The general term, an, is the last term on the left hand side. an = n2

The nth partial sum, Sn, is the right hand side. Sn = n (n + 1) (2n + 1) / 6

Find the next term in the general sequence and the series

The next term in the sequence is ak+1 and is found by replacing n with k+1 in the general term of the sequence, an.

ak+1 = ( k + 1 )2

The next term in the series is Sk+1 and is found by replacing n with k+1 in the nth partial sum, Sn. You may wish to simplify the next partial sum, Sk+1

Sk+1 = (k+1) [ (k+1) + 1 ] [ 2(k+1) + 1 ] / 6

Sk+1 = ( k + 1 ) ( k + 2 ) ( 2k +3 ) / 6 (This will be our Goal in step 3)

We will use these definitions later in the mathematical induction