A Challenge Problem in Techniques Of Problem Solving

[AK2]

I just need a hint to solve the problem. The method used is illustrated in the example problem as follows:

Assume that k is a positive integer . What is the sum of integers
S = 1+2+3+...+(k-1)+k?

Solution:

2² - 1² = 3 = 2 * 1 + 1
3² - 2² = 5 = 2 * 2 + 1
4² - 3² = 7 = 2 * 3 + 1
...
k² - (k-1)² = 2 * (k-1) + 1
(k+1)² - k² = 2 * k + 1

Now we add the columns:
[2² - 1²] + [3² - 2²] +...+ [(k+1)² - k²]
= [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 - 1² = 2[1+2+3+...k] + [1+1+1+...+1]
k times
or
k² + 2 * k = 2 * S + k

Solving for S for we find that

S = (k²+k)/2

I am supposed to imitate the method used in the example problem to solve the challenge problem in the book as follows

Imitate the method used in the last problem to find a formula for the sum
1² + 2² + 3² + ... + k²

when k is a positive integer.

I have tried various things like (a+1)⁴ - a⁴, (a+1)³ - a³, triangle numbers, odd numbers, but I haven't been able to solve it. I just need a hint. This is not a homework problem. This is self study.

 

[ross_tang]

Use

(k+1)^3 - k^3 = 3k^2+3k+1

and do summation on both side. By telescoping method, only 2 terms left in the left hand side. You need to do a little manipulation on the right hand side though.

 

[taozhen]

Why not like the Contiguous Numbers: for example,P=1+2+3+4+5...+100 and the answer of P is (1+100)*100/2 , so the S=(1+k)*k/2

 

[AK2]

Use

(k+1)^3 - k^3 = 3k^2+3k+1

and do summation on both side. By telescoping method, only 2 terms left in the left hand side. You need to do a little manipulation on the right hand side though.

Thanks for the hint
I got my answer for S

S = (2k³+3k²+k)/6

 

[ross_tang]

You are right. You may want to do some factorization also.

S = \frac{ n(n+1)(2n+1)}{6}