- #1

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I am supposed to imitate the method used in the example problem to solve the challenge problem in the book as followsAssume that k is a positive integer . What is the sum of integers

S = 1+2+3+...+(k-1)+k?

Solution:

2^{2}- 1^{2}= 3 = 2 * 1 + 1

3^{2}- 2^{2}= 5 = 2 * 2 + 1

4^{2}- 3^{2}= 7 = 2 * 3 + 1

...

k^{2}- (k-1)^{2}= 2 * (k-1) + 1

(k+1)^{2}- k^{2}= 2 * k + 1

Now we add the columns:

[2^{2}- 1^{2}] + [3^{2}- 2^{2}] +...+ [(k+1)^{2}- k^{2}]

= [2 * 1+1] + [2*2+1] + [2*3+1]+...+[2*k+1]

The left hand side ''telescopes'' (that is, all but the first and last terms cancel) and the right side may be factored. The result is

(k+1)^{2}- 1^{2}= 2[1+2+3+...k] + [1+1+1+...+1]

k times

or

k^{2}+ 2 * k = 2 * S + k

Solving for S for we find that

S = (k^{2}+k)/2

Imitate the method used in the last problem to find a formula for the sum

1

^{2}+ 2

^{2}+ 3

^{2}+ ... + k

^{2}

when k is a positive integer.

I have tried various things like (a+1)

^{4}- a

^{4}, (a+1)

^{3}- a

^{3}, triangle numbers, odd numbers, but I havent been able to solve it. I just need a hint. This is not a homework problem. This is self study.