搜尋結果
-40/27
- Constant term : 15 - 5r = 0 15 = 5 r r = 15/5 = 3 = 5C3 (-1/3)3 (2)5-3 x15 - 5 (3) = (-10/27) ⋅ 4 = -40/27 So, the constant term is -40/27.
www.onlinemath4all.com/how-to-find-the-constant-term-in-a-binomial-expansion.htmlHOW TO FIND THE CONSTANT TERM IN A BINOMIAL EXPANSION - onlinemath4all
其他人也問了
How do you find the constant term in a binomial expansion?
What is a binomial theorem?
Can A binomial theorem be generalized?
What is the constant term of (X + A)N?
What is the condition for a constant term in a binomial?
How do you find a binomial if a variable is a constant?
HOW TO FIND THE CONSTANT TERM IN A BINOMIAL EXPANSION. Example 1 : Find the constant term of (2x3 - (1/3x2))5. Solution : = (2x3 - (1/3x2))5. General term Tr+1 = nCr x(n-r) ar. x = 2x3, n = 5, a = (-1/3x2) Tr+1 = 5Cr (2x3) 5-r (-1/3x2) r. = 5Cr (2)5-r x15 - 3r (-1/3)r x-2r. = 5Cr (-1/3)r (2)5-r x15 - 5r. Constant term : 15 - 5r = 0. 15 = 5 r.
2014年9月26日 · The constant term of (x +a)n is always an; for example, the constant term of (x +3)7 is 37. Answer link. Alex Boroda · Media Owl. Jun 3, 2015. The expansion of a binomial is given by the Binomial Theorem: (x +y)n = (n 0) ⋅ xn + (n 1) ⋅ xn−1 ⋅ y1 +... +(n k) ⋅ xn−k ⋅ yk +... + (n n) ⋅ yn = n ∑ k=0 ⋅ (n k) ⋅ xn ...
In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial ( x + y ) n into a sum involving terms of the form ax b y c , where the exponents b and c are nonnegative integers with b + c = n , and the coefficient a ...
What Is the Constant Term in the Binomial Theorem? The constant term in the binomial expansion is a numeric value and is independent of the variables. For a binomial expansion of (x + y) n the term independent of x can be calculated by finding the term
- Exponents
- Exponents of
- The Pattern
- Coefficients
- As A Formula
- Putting It All Together
- Use It
- Geometry
- Isaac Newton
First, a quick summary of Exponents. An exponent says how many timesto use something in a multiplication. An exponent of 1means just to have it appear once, so we get the original value: An exponent of 0means not to use it at all, and we have only 1:
Now on to the binomial. We will use the simple binomial a+b, but it could be any binomial. Let us start with an exponent of 0and build upwards.
In the last result we got: a3 + 3a2b + 3ab2 + b3 Now, notice the exponents of a. They start at 3 and go down: 3, 2, 1, 0: Likewise the exponents of bgo upwards: 0, 1, 2, 3: If we number the terms 0 to n, we get this: Which can be brought together into this: an-kbk How about an example to see how it works:
We are missing the numbers (which are called coefficients). Let's look at all the results we got before, from (a+b)0 up to (a+b)3: And now look at just the coefficients(with a "1" where a coefficient wasn't shown): Armed with this information let us try something new ... an exponent of 4: And that is the correct answer (compare to the top of the pa...
Our next task is to write it all as a formula. We already have the exponents figured out: an-kbk But how do we write a formula for "find the coefficient from Pascal's Triangle"... ? Well, there issuch a formula: It is commonly called "n choose k" because it is how many ways to choose k elements from a set of n. The "!" means "factorial", for exampl...
The last step is to put all the terms together into one formula. But we are adding lots of terms together ... can that be done using one formula? Yes! The handy Sigma Notationallows us to sum up as many terms as we want: Sigma Notation Now it can all go into one formula:
OK ... it won't make much sense without an example. So let's try using it for n = 3: BUT ... it is usually much easier just to remember the patterns: 1. The first term's exponents start at n and go down 2. The second term's exponents start at 0 and go up 3. Coefficients are from Pascal's Triangle, or by calculation using n!k!(n-k)! Like this: We ma...
The Binomial Theorem can be shown using Geometry: In 2 dimensions, (a+b)2 = a2 + 2ab + b2 In 3 dimensions, (a+b)3 = a3 + 3a2b + 3ab2 + b3 In 4 dimensions, (a+b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 (Sorry, I am not good at drawing in 4 dimensions!)
As a footnote it is worth mentioning that around 1665 Sir Isaac Newton came up with a "general" version of the formula that is not limited to exponents of 0, 1, 2, .... I hope to write about that one day.
Using the Binomial Theorem to Find a Single Term Expanding a binomial with a high exponent such as (x + 2 y) 16 (x + 2 y) 16 can be a lengthy process. Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully (x + y
Sometimes it is helpful to identify the pattern that results from applying the binomial theorem. Notice that powers of the variable \(x\) start at \(5\) and decrease to zero. The powers of the constant term start at \(0\) and increase to \(5\). The binomial coefficients
