Sum to n terms of the series 1+5+12+22..........

5[x(2+y)] What is the associative law to of addition to this expression? Wrong forum category but... The associative law of addition does just that: it associates some factor (some multiplier) in an expression with all of the terms in string of addends by which the factor will be multiplies. I na nutshell, the laws holds, for example, that in the expression, 5[x(2+y)], the factor, x, can be multiplied by each term of the sum (2+y), without first finding their total, to arrive at (2x + yx); and, in turn, those terms can be separately associated again with the factor, 5, in order to arrive at the equation,
5(2x + yx) = 10x + 5xy. Note also that the law includes its reverse, by which terms can be dissociated and summed prior to being multiplied by a former distributed factor.

If its x2 over x2, the solution is 1. x2/x2=1
please contact me for more help if this is not what you mean.

What kind of sum would that be? If it is a sum of a certain number of values that are explicitly given (2+5.5+ 16...) just enter them one after the other separated by the operator (+).
For a sum of terms that are defined by some recursion fornula then you will need a calculator with a Computer Algebra System (TI Nspire CAS, TI 92PLUS, TI Voyage 200 PLT, Casio ClassPad, HP 40G/49 G+/50G..)
A TI 83PLUS/84PLUS may be able to handle numerical sums of terms defined recursively.

Here, We deal with Some Special Products in Polynomials.

Certain products of Polynomials occur more often
in Algebra. They are to be considered specially.

These are to be remembered as Formulas in Algebra.

Remembering these formulas in Algebra is as important
as remembering multiplication tables in Arithmetic.

We give a list of these Formulas and Apply
them to solve a Number of problems.

We give Links to other Formulas in Algebra.

Here is the list of Formulas in
Polynomials which are very useful in Algebra.
Formulas in Polynomials :

Algebra Formula 1 in Polynomials:

Square of Sum of Two Terms:

(a + b)2 = a2 + 2ab + b2
Algebra Formula 2 in Polynomials:

Square of Difference of Two Terms:

(a - b)2 = a2 - 2ab + b2
Algebra Formula 3 in Polynomials:

Product of Sum and Difference of Two Terms:

(a + b)(a - b) = a2 - b2
Algebra Formula 4 in Polynomials:

Product giving Sum of Two Cubes:

(a + b)(a2 - ab + b2) = a3 + b3
Algebra Formula 5 in Polynomials:

Cube of Difference of Two Terms:

(a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - 3ab(a - b) - b3
Algebra Formula 8 in Polynomials:


Each of the letters in fact represent a TERM.

e.g. The above Formula 1 can be stated as
(First term + Second term)2
= (First term)2 + 2(First term)(Second term) + (Second term)2

This sum is equal to 40+48+56+64+72+...+3024+3032+3040+3048= 40+(48+3048)+(56+3040)+(64+3032)+(72+3024)...
There are (3048-40)/8+1=377 terms in sum. (48+3048)=(56+3040)=(64+3032)=(72+3024)=3096 and there are (377-1)/2=188 pairs whose sum is 3096. Only term that is not in such pair is 40.
Finally sum is 40+188*3096=582088.

Compare the terms of the product of the square of binomial and the terms of the product of the sum and difference of two terms. What statements can you make?

Using elementary algebria in the binomial theorem, I expanded the power (x + y)^n into a sum involving terms in the form a x^b y^c. The coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n. This is known as binomial coefficients and are none other than combinatorial numbers.

Combinatorial interpretation:

Using binomial coefficient (n over k) allowed me to choose k elements from an n-element set. This you will see in my calculations on my Ti 89. This also allowed me to use (x+y)^n to rewrite as a product. Then I was able to combine like terms to solve for the solution as shown below.
(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y) = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

This also follows Newton's generalized binomial theorem:



Now to solve using the Ti 89.

Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:

The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n ? k) + k, always equals n.






The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.





This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.




This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y






This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2







This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3







This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4






This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5





This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6





Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.

binomial coefficient (n over k) for (x+y)^6
x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6

The sum of the first counting numbers 1 + 2 + 3 + 4 + 5 +... + is given by
Use , where is the first term in the series, and is the last.

Now to solve this for nos, 1 to 10000 inclusive

Subtract 10000 - 1 = 9999 and add 1 to get 10000 terms in the series.

S = (10000[ 1+10000])/2 = 50,005,000.

This should do it.

hahahhaha.. because a^b + c^b is differnet from (a+c)^b... factorization... dude.. factorization!!