Question 40302: Hello!
Here's a tutor who is a bit rusty on this subject. Is there a simple expression for the following sum?
For example,
1 + 4 + 9 + 16 + ... + 100
or
25 + 36 + 49
etc.
Thank you very much!
Found 2 solutions by venugopalramana, AnlytcPhil:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Here's a tutor who is a bit rusty on this subject. Is there a simple expression for the following sum?
For example,
1 + 4 + 9 + 16 + ... + 100....we find these are all squares of 1,2,3 etc..
so we say
TN=N^2...PUT N=1,2,3, ..ETC...TO GET SUCCESSIVE TERMS..1^2=1,2^2=4,3^2=9....ETC
TO SHOW SUM WE SAY FIND SN FOR N=10..THAT IS FROM N=1 TO 10..THAT IS WHAT YOU PUT ABOVE
OFCOURSE IT IS ALSO REPRESENTED BY WHAT YOU GAVE IN THE BEGINING NAMELY
SIGMA(N^2)...N RANGING FROM 1 TO 10 OR WHATEVER..FOR THE EXAMPLE YOU GAVE
1+4+9+16+.....+100...TO PUT A TO A+(N-1)IS NOT CORRECT.YOU SHOULD PUT ONLY N=1 TO N=10
or
25 + 36 + 49
etc.
Thank you very much!
Answer by AnlytcPhil(1807)  (Show Source):
You can put this solution on YOUR website! I'm rusty too, so I'll have to derive it from scratch.
First we'll find a formula for
S(k) = 
then your summation is
S(a+n-1) - S(a-1)
{S(k)} = 1², 1²+2², 1²+2²+3², ··· , 1²+2²+3²+···+k²
Write out some of the terms
1, 5, 14, 30, 55, 91, 140
Make a difference table, i.e, list the terms
vertically in a column placing the difference
between each pair of successive terms between
them to the right. Then do the same to the
second column, until you get a column which all
contain the same number:
1
4
5 5
9 2
14 7
16 2
30 9
25 2
55 11
36 2
91 13
49
140
It took three difference columns to get a column of
all 2's, so we will see if a third degree polynomial
in k is possible for the formula:
S(k) = Ak³ + Bk² + Ck + D
Then substituting
1 = A(1)³ + B(1)² + C(1) + D
5 = A(2)³ + B(2)² + C(2) + D
14 = A(3)³ + B(3)² + C(3) + D
30 = A(4)³ + B(4)² + C(4) + D
Giving us this system of 4 equations in 4 unknowns
A + B + C + D = 1
8A + 4B + 2C + D = 5
27A + 9B + 3C + D = 14
64A + 16B + 4C + D = 30
Solve this system and we get
A = 1/3, B = 1/2, C = 1/6, D = 0
So if we are right our formula is
S(k) = (1/3)k³ + (1/2)k² + (1/6)k
S(k) = (2k³ + 3k² + k)/6
S(k) = k(2k² + 3k + 1)/6
S(k) = k(k+1)(2k+1)/6
This can be proved correct by induction.
It is true for k=1, so if we add (k+1)²
to both sides
S(k) + (k+1)² = k(k+1)(2k+1)/6 + (k+1)²
= [(k+1)/6][k(2k+1) + 6(k+1)]
= [(k+1)/6][2k²+7k+6] = [(k+1)/6][(k+2)(2k+3)]
= (k+1)(k+2)(2k+3)/6 which equals S(k+1)
so we have the right formula.
Now your problem is to find
S(a+n-1) - S(a-1)
(a+n-1)(a+n-1+1)(2(a+n-1)+1)/6 -
(a-1)(a-1+1)(2(a-1)+1)/6
(a+n-1)(a+n)(2a+2n-1)/6 - (a-1)(a)(2a-1)/6 =
[(a+n-1)(a+n)(2a+2n-1) - a(a-1)(2a-1)]/6
I'll let you simplify that if you like.
I'm too tired! :-)
Edwin
AnlytcPhil@aol.com
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