Expert: Sherry Wallin - 5/13/2009

Question
QUESTION: Sir, I just read a solution of the following question.
What is the sum of 1st n odd integers..?

And the solution was as follows.

Look at each of the terms.
For n=1, the sum is 1.
For n=3, the sum is 1+9=10.
For n=5, the sum is 1+9+25=35.
For n=7, the sum is 35+49=84.

If a difference table is used, you get
1   1
3  10  9/2
5  35 25/2  2  
7  84 49/2  3  1
9 165 81/2  4  1  0

If you need to know how to construct a difference table, here's how.
9/2=(10-1)/(3-1); 25/2=(35-10)/(5-3); 2=(25/2 - 9/2)/(5-1); etc.

We then need to work out the formula
[{1(x-5) + 2}(x-3) + 4.5](x-1) + 1.  This will give you the equation passing through these points.  To start off, note that the leading 1 can be dropped, lettering us take (x-5)+2=(x-3).  Then take that (x-3)(x-3) and add on 4.5.  Once this has been done, multiply by (x-1) and add 1.

I wanna ask that this solution is working up to x=3,5

While I took x=7 then the answer must have been 84.But I cant get it.
According to the solution, I did " (7-3)(7-3)=4*4=16+4.5=20.5*(7-1)=123+1=124 that is not equal to 84....
So, how can we derive it?


ANSWER: Hi Aniket~
    Something is wrong with the data you were given. The first three odd integers is 1+3+5 = 9 not 10 and the first 5 is 9+7+9 = 25 not 35 and the first 7 is 25+11+13 = 49 not 84. These beginning errors propagate the error you are experiencing.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Hello sir, thanks for your kind reply,

I mistakenly typed a wrong question earlier.
My Que was " What is the sum of square of 1st n odd nos?". In that case that whole working is applicable.

1^2 +3^2= 10
1^2 +3^2 +5^2= 35
1^2 +3^2 +5^2 +7^2 =84

And i am not able to get the answer 84 with the help of that solution... So, kindly solve my problem sir....

ANSWER: Hi Aniket~
    I think there is still a problem with what you gave me. In your difference table your third entry is incorrect: 2=(25/2 - 9/2)/(5-1) for two reasons, I believe the first 2 is suppose to be (?)/2 instead of 2 and maybe then I can tell you if  the 5-1 is correct or not.

Math Prof

The difference table I come up with for the first 5 summands is:
1
   9
10     34
   25    10
35     24    2
   49     8
84     32
   81
165

---------- FOLLOW-UP ----------

QUESTION: Thanks for your reply again.

I think It ll be complicated if I try solving this sum with the help of Difference table... So, is there any other way to solve it??

Like by using Geometric Progression???
terms ll be (1*1)+(3*3)+(5*5)............. and so on.

So, Is there any way we can solve this by GP?
Kindly reply soon.

Answer
Aniket~
    Thank you for your patience. It still is not clear what you mean by 'solve it'? To solve means it must be an equation. Students often misuse the word solve when they mean simplify. We simplify expressions and solve equations. If I hazard a guess I think you mean you want to find the pattern and rewrite it compactly. Let's approach this from a slightly different direction. I am going to use capital S to mean the Summation Greek symbol and I will use (x,y) to mean the range I will sum over and all odd integers 'look like' 2k+1 so the problem becomes S(k = 0, n)(2k-1)^2 which will generate 2(1)-1 + 2(2)-1 + 2(3)-1 +...+2(n) -1. Are you with me to here?
Now (2k-1)^2 is really 4k^2-4k+1 (expanded) and if I sum it from k=1 to n then k runs over n  positions (odd numbers). So the sum of the first n squared odd integers is just n(4k^2-4k+1). Is this what you are looking for?

Math Prof