Write in sigma notation

[mollymvora]

2 + 5 + 10 + 17 + 26 + 37
write that in sigma notation.

 

[Deleted member 4993]

2 + 5 + 10 + 17 + 26 + 37
write that in sigma notation.

First check the differences between terms:

2         5          10         17          26          37 .............

     3          5           7            9            11  ................

           2           2           2            2  .....................

                 0          0            0       ..........................

so the terms above follow afunction of second order of the type

a[sub:tdf27xzl]n[/sub:tdf27xzl] = A[sub:tdf27xzl]0[/sub:tdf27xzl] + A[sub:tdf27xzl]1[/sub:tdf27xzl] * n + A[sub:tdf27xzl]2[/sub:tdf27xzl] * n[sup:tdf27xzl]2[/sup:tdf27xzl]

Calculate the values of A[sub:tdf27xzl]0[/sub:tdf27xzl] , A[sub:tdf27xzl]1[/sub:tdf27xzl] & A[sub:tdf27xzl]2[/sub:tdf27xzl] from the given data - you could probably come up with the expression by some judicious observation.

Then write:

\(\displaystyle S_6 \ = \ \sum_{n=1}^{n=6} \left [A_0 + A_1* n + A_2 * n^2\right ]\)

 

[mollymvora]

2 + 5 + 10 + 17 + 26 + 37
write that in sigma notation
Why does it have to be that complicated? because the pattern for the differences are 2n + 1 so is there any way of doing this in a less complicated manner?

 

[soroban]

Hello, mollymvora!

I "eyeballed" it . . .

\(\displaystyle \text{Write in sigma notation: }\:2 + 5 + 10 + 17 + 26 + 37\)

\(\displaystyle \text{I saw that each term is one more than a square:}\)

. . . . \(\displaystyle \begin{array}{ccc} 2 &=& 1^2 + 1 \\ 5 &=& 2^2+1 \\ 10 &=& 3^2 + 1 \\ 17 &=& 4^2+1 \\ 26 &=& 5^2+1 \\ 37 &=& 6^2+1 \end{array}\)

\(\displaystyle \text{So the }n^{th}\text{ term is: }\:n^2+1\)

. . \(\displaystyle \text{Answer: }\;\sum^6_{n=1}(n^2+1)\)

 

[Deleted member 4993]

2 + 5 + 10 + 17 + 26 + 37
write that in sigma notation
Why does it have to be that complicated? because the pattern for the differences are 2n + 1 so is there any way of doing this in a less complicated manner?

Since you did not show any work - I showed you the most general (and fail-proof) way to solve the problem.

If you have done thousands of these types of problems - then you could eye-ball the solution, like Soroban did (then ofcourse you would not be asking for help).

To review the mathematical methods for such problems, go to:

http://www.purplemath.com/modules/nextnumb.htm

For more advanced interpretation study up on (Gregory-Newton Formula) at:

http://mathworld.wolfram.com/NewtonsFor ... rmula.html