Find
1. the binomial expansion of (3+2x)^5 simplifying terms. Hence find the binomial expansion of (3+2x)^5 +(3-2x)^5.
(3+2x)^5
= 3^5 + 5(3^4)(2x) + 10(3^3)(2x)^2 + 10(3^2)(2x)^3 + 5(3)(2x)^4 + (2x)^5
=243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5
by simple observation that a negative base raised to an odd exponent is negative ....
(3-2x)^5
= 243 - 810x + 1080x^2 - 720x^3 + 240x^4 - 32x^5
(3+2x)^5 +(3-2x)^5
= 243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5 + 243 - 810x + 1080x^2 - 720x^3 + 240x^4 - 32x^5
= 486 + 2160x^2 + 480x^4
confirmed by Wolfram
http://www.wolframalpha.com/input/?i=expand+%283%2B2x%29%5E5+%2B%283-2x%29%5E5
Oh, binomial expansion, you've always been a party animal. Let's get into it!
The binomial expansion of (3 + 2x)^5 can be found using the binomial theorem. Here's how it goes:
(3 + 2x)^5 = (5C0)*(3^5)*(2x)^0 + (5C1)*(3^4)*(2x)^1 + (5C2)*(3^3)*(2x)^2 + (5C3)*(3^2)*(2x)^3 + (5C4)*(3^1)*(2x)^4 + (5C5)*(3^0)*(2x)^5
Now let me simplify those terms for you, my dear friend:
(1)*(243)*(1) + (5)*(81)*(2x) + (10)*(27)*(4x^2) + (10)*(9)*(8x^3) + (5)*(3)*(16x^4) + (1)*(1)*(32x^5)
Simplifying that mess, we get:
243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5
Now let's move on to the next part of your question. We need to find the binomial expansion of (3+2x)^5 + (3-2x)^5. But before we begin, I must warn you that adding two binomial expansions can sometimes lead to unexpected results, just like mixing peanut butter and pickles. So, brace yourself!
The first thing we need to do is find the binomial expansion of (3-2x)^5, which follows the same procedure as before:
(3 - 2x)^5 = (5C0)*(3^5)*(-2x)^0 + (5C1)*(3^4)*(-2x)^1 + (5C2)*(3^3)*(-2x)^2 + (5C3)*(3^2)*(-2x)^3 + (5C4)*(3^1)*(-2x)^4 + (5C5)*(3^0)*(-2x)^5
Simplifying those terms with my hilarious touch, we have:
(1)*(243)*(1) + (-5)*(81)*(2x) + (10)*(27)*(4x^2) + (-10)*(9)*(8x^3) + (5)*(3)*(16x^4) + (-1)*(1)*(32x^5)
Making sense of that crazy formula, we get:
243 - 810x + 1080x^2 - 720x^3 + 240x^4 - 32x^5
Now, let's add those two expansions together, like trying to teach a clown to juggle:
(243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5) + (243 - 810x + 1080x^2 - 720x^3 + 240x^4 - 32x^5)
Drumroll, please... Simplifying that fusion of binomial goodness, we get:
486 + 2160x^2 + 480x^4
And there you have it, my friend! The binomial expansion of (3+2x)^5 + (3-2x)^5 is 486 + 2160x^2 + 480x^4. I hope this mathematical circus act brought a smile to your face!
To find the binomial expansion of (3+2x)^5, we can use the binomial theorem. The binomial theorem states that for any positive integer n:
(a + b)^n = C(n,0)a^n b^0 + C(n,1)a^(n-1) b^1 + C(n,2)a^(n-2) b^2 + ... + C(n,n-1)a^1 b^(n-1) + C(n,n)a^0 b^n
Where C(n,k) is the binomial coefficient, given by:
C(n,k) = n! / (k!(n-k)!)
Let's find the binomial expansion of (3+2x)^5:
Using the formula, we have:
(3+2x)^5 = C(5,0)(3)^5 (2x)^0 + C(5,1)(3)^4 (2x)^1 + C(5,2)(3)^3 (2x)^2 + C(5,3)(3)^2 (2x)^3 + C(5,4)(3)^1 (2x)^4 + C(5,5)(3)^0 (2x)^5
Simplifying terms, we get:
(3+2x)^5 = 1(3)^5 + 5(3)^4 (2x) + 10(3)^3 (2x)^2 + 10(3)^2 (2x)^3 + 5(3)^1 (2x)^4 + 1(2x)^5
This simplifies to:
(3+2x)^5 = 243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5
Now, let's find the binomial expansion of (3+2x)^5 +(3-2x)^5:
Using the binomial theorem, we can find the expansion of (3-2x)^5 similarly to what we did above. Then, we can add the corresponding terms from the expansions of (3+2x)^5 and (3-2x)^5.
(3+2x)^5 = 243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5
(3-2x)^5 = 243 - 810x + 1080x^2 - 720x^3 + 240x^4 - 32x^5
Adding the corresponding terms, we get:
(3+2x)^5 +(3-2x)^5 = (243 + 243) + (810x - 810x) + (1080x^2 + 1080x^2) + (720x^3 - 720x^3) + (240x^4 + 240x^4) + (32x^5 - 32x^5)
Simplifying, we get:
(3+2x)^5 +(3-2x)^5 = 486 + 2160x^2 + 480x^4
Therefore, the binomial expansion of (3+2x)^5 +(3-2x)^5 is 486 + 2160x^2 + 480x^4.
To find the binomial expansion of (3+2x)^5, we can use the binomial theorem. The binomial theorem states that (a + b)^n can be expanded as:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
where C(n, k) represents the binomial coefficient, given by:
C(n, k) = n! / (k! * (n-k)!), where n! represents the factorial of n.
Now let's find the binomial expansion of (3+2x)^5 by substituting the values into the binomial theorem formula:
(3+2x)^5 = C(5, 0) * 3^5 * (2x)^0 + C(5, 1) * 3^4 * (2x)^1 + C(5, 2) * 3^3 * (2x)^2 + C(5, 3) * 3^2 * (2x)^3 + C(5, 4) * 3^1 * (2x)^4 + C(5, 5) * 3^0 * (2x)^5
Simplifying each term, we get:
(3+2x)^5 = 1 * 3^5 * 1 + 5 * 3^4 * 2x + 10 * 3^3 * (2x)^2 + 10 * 3^2 * (2x)^3 + 5 * 3^1 * (2x)^4 + 1 * 3^0 * (2x)^5
Now we can find the binomial expansion of (3+2x)^5 + (3-2x)^5 by simply adding the corresponding terms of the two expansions:
(3+2x)^5 + (3-2x)^5 = (1 * 3^5 * 1 + 5 * 3^4 * 2x + 10 * 3^3 * (2x)^2 + 10 * 3^2 * (2x)^3 + 5 * 3^1 * (2x)^4 + 1 * 3^0 * (2x)^5) + (1 * 3^5 * 1 - 5 * 3^4 * 2x + 10 * 3^3 * (2x)^2 - 10 * 3^2 * (2x)^3 + 5 * 3^1 * (2x)^4 - 1 * 3^0 * (2x)^5)
Simplifying the terms inside the parentheses, we get:
(3+2x)^5 + (3-2x)^5 = (243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5) + (243 - 810x + 1080x^2 - 720x^3 + 240x^4 - 32x^5)
Now we can combine like terms:
(3+2x)^5 + (3-2x)^5 = 2 * 243 + 2 * 810x + 2 * 1080x^2 + 2 * 720x^3 + 2 * 240x^4
Simplifying further:
(3+2x)^5 + (3-2x)^5 = 486 + 1620x + 2160x^2 + 1440x^3 + 480x^4
Therefore, the binomial expansion of (3+2x)^5 + (3-2x)^5 is 486 + 1620x + 2160x^2 + 1440x^3 + 480x^4.