Algebra 2

Quote:

The problem is: (3t + 4)^5
Instead of simply expanding, my instructor wants me to use Pascal's Triangle.

Neat question.

(a+b) ⁿ can be written as

Sum[k={0,n} Choose(n,k) a ^kb ^n-k]

i.e. Choose(n,0) b ⁿ+Choose(n,1) a b ^n-1 + ... Choose(n,n-1) a ^n-1b + Choose(n,n) a ⁿ

Now you're probably thinking ok wth is "Choose"?

This is the binomial coefficient, "n choose k" and they are the values of a given row in Pascal's triangle. Come down n rows and go over k elements in the triangle and you get the Binomial coefficient corresponding to "n choose k" = Choose(n,k) = Binomial(n,k) (different names for the same thing).

So in this case if you look at the n=5 row (really the 6th row, the first row is n=0) of Pascal's triangle you see it's

1 5 10 10 5 1

in other words

Choose(5,0)=Choose(5,5) = 1
Choose(5,1)=Choose(5,4) = 5
Choose(5,2)=Choose(5,3) = 10

so (a+b) ⁵ = 1*a ⁵ + 5 a ⁴ b + 10 a ³ b ² + 10 a ² b ³ + 5 a b ⁴ + 1 b ⁵

now let a=3t and b=4 and substitute those into the formula above and simplify things and you're done

This is what I think it is although I learned this 2 months ago
(a+b)^x

1(3t)^5+5(3t)^4(4)^1+10(3t)^3(4)^2+10(3t)^2(4)^3+5(a)^1(b)^4+1(b)^5

the numbers 1,5,10,10,5,1 before each set of a's and/or b's are part of the triangle. If the equation were to the ^4th then it would be 1,4,6,4,1

your final answer though is:

243t^5+1604t^4+4320t^3+5760t^2+3840t+1024

Thanks for the quick responses! First time using this site and I'll certainly come back.

Ethan Bates:

The problem is: (3t + 4)^5
Instead of simply expanding, my instructor wants me to use Pascal's Triangle.

You have alreay had 2 good answers, I am offering a third.

140409 - Pascal's triangle.JPG

This is pascal's triangle - you are going to use the row where the 5 is the second number.

(3t + 4) ⁵= 1*(3t) ⁵*4 ⁰ + 5* (3t) ⁴*4 ¹ + 10*(3t) ³*4 ² + 10*(3t) ²*4 ³ + 5*(3t) ¹*4 ⁴ + 1*(3t) ⁰*4 ⁵

=(3t) ⁵ + 5* (3t) ⁴*4 + 10*(3t) ³*4 ² + 10*(3t) ²*4 ³ + 10*(3t)*4 ⁴ + 4 ⁵