Math Help

Archive for September, 2008

Question: How do I show by mathematical induction that: the sumation from r=1 to n of r^2 is equal to n/6(n+1)(2n+1)?

∑_r=1 to n (r^2) = n/6(n+1)(2n+1)

Answer: Proof by induction, first check it works for the base case
when n=1
which gives
∑r=1 to n (r^2)=1
and also
n/6(n+1)(2n+1) =1

so for the case of n=1
∑r=1 to n (r^2) = n/6(n+1)(2n+1)
is a true statement.

now we assume the statement is true for the case k. so we have when n=k

∑r=1 to k (r^2) = k/6(k+1)(2k+1)
is true. now we need to show that for the case k+1 the statement is also true if it is you have completed proof by induction.

so
∑r=1 to k+1 (r^2) = (k+1)^2 + ∑r=1 to k (r^2)

all I’m saying here is the sum for 1 to k+1 is the same as the sum as 1 to k plus
(k+1)^2, (it can be difficult to decipher when written on keyboard instead of by hand, so I’ve explained it here). So when we use the assumption of the case k is true and subtitute in (k/6)(k+1)(2k+1) for ∑r=1 to k (r^2) we get:

∑r=1 to k+1 (r^2) = (k+1)^2 + (k/6)(k+1)(2k+1)

now use loooking at the right hand side you will need to expand the bractkets adn simplfy down as given below

(k+1)^2 + (k/6)(k+1)(2k+1) expand to get
k^2+2k+1 + (k/6)(2k^2+3k+1)
k^2+2k+1 + (1/6)(2k^3+3k^2+k) first bit multiply by 6/6
(1/6)(6k^2+12k+6) + (1/6)(2k^3+3k^2+k) now 1/6 can factorise out
1/6[6k^2+12k+6 + 2k^3+3k^2+k] simplify
1/6[2k^2 + 9k^2 + 13k + 6] factoise out the term k+1
[(k+1)/6](2k^2+7k+6) and factorise again to get
[(k+1)/6](k+2)(2k+3)
[(k+1)/6][(k+1)+1][2(k+1)+1]

thus for the case of n = k+1 we get
∑r=1 to n (r^2) = [(k+1)/6][(k+1)+1][2(k+1)+1]
or as n=k+1
∑r=1 to n (r^2) = (n/6)(n+1)(2n+1)

as required

1, 2, 3 … Infinity. Mathematical Induction

Question: Math Problem solve pls…?

SEND
+MORE
=MONEY

Solve For: S, E, N, D, M, O, R, Y
pls tell me, is there a MATH Software that can help me solve these problem?

Answer: That’s actually pretty nice problem.

The solution is: S = 9, E = 5, N = 6, D = 7, M = 1, O = 0, R = 8, Y = 2.
Therefore:
9567
+1085
=10652

There probably is also some math software to solve that kind of problems, but it really wasn’t that hard to solve using just a pen and some paper.

Algebra Software – YourTeacher.com – 1000+ Online Math Lessons

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Shown in the picture is a 3 by 2 matrix. Horizontal lines of matrices are called rows while vertical lines are called columns. This matrix has 3 rows and 2 columns so it is a 3 by 2 matrix. It can also be written as 3 x 2 matrix. 3 and 2 are called the dimension of the matrix. Dimension of matrices are given by the number of rows of the matrices first then the number of columns. Matrices can have any dimensions.

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