A Challenge Problem in Techniques Of Problem Solving

  • Thread starter AK2
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  • #1
AK2
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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 havent been able to solve it. I just need a hint. This is not a homework problem. This is self study.
 

Answers and Replies

  • #2
64
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Use

[tex] (k+1)^3 - k^3 = 3k^2+3k+1[/tex]

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.
 
  • #3
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
 
  • #4
AK2
39
0
Use

[tex] (k+1)^3 - k^3 = 3k^2+3k+1[/tex]

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
 
  • #5
64
0
You are right. You may want to do some factorization also.

[tex] S = \frac{ n(n+1)(2n+1)}{6}[/tex]
 

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