What is the binomial expansion of [math](1+x) ^{-2}[/math]?
Hope this will help to build fundamentals of binomial..
What is the binomial expansion for (1+x)^-1?
Hello,Binomial expansion for this is followed the same way as expanding larger power expansionsSo (1+x)^-1 = 1 + (-1)x + (-1)(-1–1)/2! x^2 + (-1)(-1–1)(-1–2)/3! x^3 +…….So simplifying we get1 -x + x^2 - x^3 +….This followes a simple formula given by(1 + x)^n = 1 + nx + n(n-1)/2!. x^2 + n(n-1)(n-2)/3!. x^3 +…..Bear in mind this will result in an infinite series and will converge for values belonging to |x| < 1
What is the binomial expansion of [math]{(1-x)}^{-n}[/math] and [math]{(1+x)}^{n}[/math]?
we know, coefficient of[math] x^{r} [/math]in expansion of [math](1+x)^{-n} [/math]is [math] ^{n+r-1}C_r[/math]therefore [math](1+x)^{-n}=\sum\limits_{r=0}^{\infty}[/math] [math]{^{n+r-1}C_r.x^{r}}[/math]similarly coefficient of[math] x^{r} [/math]in expansion of [math](1+x)^{n} [/math]is [math]^{n}C_r [/math]therefore[math] (1+x)^{n}=\sum\limits_{r=0}^{n}[/math] [math]^{n}C_r. x^{r}[/math]P.S. [math](1+x)^{-n}[/math] contains infinite terms in its expansion whereas [math](1+x)^{n}[/math] contains[math] {n+1}[/math] terms
Binomial expansion help?
Assuming, you are not using a calculator, it will be easier to work with integers rather than with fractions/decimals. So multiply by a form of 1 to get rid of the decimal: 100^5 * 1.04^5 / 100^5 = (100 * 1.04)^5 / 100^5 = 104^5 / 100^5 We will divide after we expand using binomial expansion. So, now we want to expand 104^5. We break it up as a sum of integers: (104)^5 = (100 + 4)^5 For some x, y, the binomial expansion is given by: (x + y)^n = Sum[ {n,k} * x^(n-k) * y^k ] for k = 0 to n where {n,k} = "n choose k" = n! / [ k! * (n - k)! ] This gives us: {5,0} * 100^5 * 4^0 + {5,1} * 100^4 * 4^1 + {5,2} * 100^3 * 4^2 + {5,3} * 100^2 * 4^3 + {5,4} * 100^1 * 4^4 + {5,5} * 100^0 * 4^5 = 1*100^5 + 5 * 100^4 * 4 + 10 * 100^3 * 16 + 10 * 100^2 * 64 + 5 * 100 * 256 + 1 * 100^0 * 4^5 = 100^5 + 20 * 100^4 + 160 * 100^3 + 640 * 100^2 + 500 * 256 + 1024 Now divide by 100^5: (100^5 + 20 * 100^4 + 160 * 100^3 + 640 * 100^2 + 500 * 256 +1024) / 100^5 = 1 + 20/100 + 160/100^2 + 640/100^3 + 1280/100^4 + 1024/100^5 = 1 + 0.2 + 0.016 + 0.00064 + 0.0000128 + 0.0000001024= 1.2166529024 Hope that helps.
Binomial expansion - please help (AS Level)?
Hi, I am trying to do a maths paper, and I'm totally stuck on this question. I'm so angry at myself because I know it should be really easy, but I just can't remember how to do it. I have the Answer Booklet, with the answer and the steps to it, but it doesn;t explain it so I don't understand. please can you do a step by step answer, and explain why you did each step, I'll be really grateful, thank you! a) Find the first 3 terms, in ascending powers of x, of the binomial expansion of (1 + px)^9 where p is a constant. b) The first 3 terms are 1, 36x and qx^2, where q is a constant. Find the value of p and q.
Binomial expansion help needed! 10pts?
For b), I'm assuming you mean 601^4. Note that 601 = 1 + 6(100), so substitute x = 100 into your binomial expansion in a) to calculate 601^4 without a calculator. Lord bless you today!
Binomial expansion[HELP PLEASE]?
(1+x+x²)^n = [1+ (x+x²)]^n = [1+ x·(1+x)]^n = 1^n+ nC1·x·(1+x)+ nC2·x²(1+x)² + nC3·x³·(1+x)³ + nC4· x^4·(1+x)^4 + nC5·x^5·(1+x)^5 +...... = 1 + n·x + [nC1+nC2] x² + [2·nC2+nC3] x³ + [nC2+3·nC3+nC4) x^4 + [ 3·nC3+4·nC4+nC5]·x^5 +..... ==> n· [ 3·nC3+4·nC4+nC5] = [nC1+nC2] · [2·nC2+nC3] n· [ 3· (nC3+nC4) + (nC4+nC5) ] = [nC1+nC2] · [nC2+(nC2+nC3)] n· [ 3· (n+1)C4 + (n+1)C5 ] = (n+1)C2 · [ nC2+ (n+1)C3] n· [ 3· (n+1)C4 + (n-3)/5 · (n+1)C4] = (n+1)C2 ·[ nC2+ (n+1)/3· nC2] n· (n+1)C4· [ 3+ (n-3)/5] = (n+1)C2· nC2 · [ 1+ (n+1)/3] n· (n+1)C4 · (n+12)/5 = (n+1)C2 · nC2 · (n+4)/3 n· (n+1)·n·(n-1)·(n-2)/24 · (n+12)/5 = (n+1)·n/2 · n·(n-1)/2 · (n+4)/3 (n-2)·(n+12)/120 = (n+4)/12 (n-2)·(n+12)= 10·(n+4) n²+ 10n - 24 = 10n + 40 n²=64 --> n= -8 or n=8 ==> Solution n=8 Saludos
How to use binomial expansions to find 0.98^8 ?
lets say the binomial expansion is (1 - 2x) ^ 8 and you want to find 0.98^8 using the first four terms in the expansions, how could you do it? Appreciate any help !!
How can I calculate the binomial expansion for (1+x) ^-n?
If n is positive integer or positive rational number, we can have the binomial expansion of [math]\;\;(1+x)^{-n}\;\;[/math] as an infinite series as follows:[math]\;\;[/math] [math]\;\;(1+x)^{-n}\;=\;1\;+\;\frac{-n}{1}.x^{1}\;+\;\frac{(-n)(-n-1)}{1 \times 2}.x^{2}\;+\;\frac{(-n)(-n-1)(-n-2) }{1 \times 2 \times 3}.x^{3}\;+\;...+\;\frac{(-n)(-n-1)(-n-2)...(-n-r+1) }{1 \times 2\;...\times r}.x^{r}\;+\;...\infty\;\;[/math]This series converges to some real/complex value, for each [math]\;x\; [/math]with [math]\;\;|x|\;<\;1\;\; [/math]For example [math]\;\;[/math] [math]\;\;(1+x)^{-1}\;=\;1\;+\;\frac{-1}{1}.x^{1}\;+\;\frac{(-1)(-1-1)}{1 \times 2}.x^{2}\;+\;\frac{(-1)(-1-1)(-1-2) }{1 \times 2 \times 3}.x^{3}\;+\;...\;+\;...\infty\;\;[/math][math]\;\;=\;\;1\;-\;x^{1}\;+\;x^{2}\;-\;x^{3}\;+\;...\infty\;\;[/math][math]\;\;[/math] [math]\;\;(1+x)^{-2}\;=\;1\;+\;\frac{-2}{1}.x^{1}\;+\;\frac{(-2)(-2-1)}{1 \times 2}.x^{2}\;+\;\frac{(-2)(-2-1)(-2-2) }{1 \times 2 \times 3}.x^{3}\;+\;...\;+\;...\infty\;\;[/math][math]\;\;=\;\;1\;-\;2x^{1}\;+\;3x^{2}\;-\;4x^{3}\;+\;...\infty\;\;[/math][math]\;\;[/math] [math]\;\;(1+x)^{-\frac{1}{2}}\;=\;1\;+\;\frac{-\frac{1}{2}}{1}.x^{1}\;+\;\frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{1 \times 2}.x^{2}\;+\;...\infty\;\;[/math][math]\;\;=\;\;1\;-\;\frac{1} {2 } x^{1}\;+\;\frac {1 \times 3} {2\times 4} x^{2}\;-\;\frac {1\times 3 \times 5} {2 \times 4\times 6} x^{3}\;+\;...\infty\;\;[/math]Actually these are Taylor series of the given function about the point zero.
Need help with Binomial expansion urgently!?
could someone please help me with the following question (exam style question),and explain in full how to do it? thanks in advance. For the binomial expansion, in descending powers of x, of (x^3 - 1/2x)^12 ( just to clarify the '2' is the coefficient of x in this case if it seemed unclear). (1) find the first 4 terms simplifying each term. (2) find in its simplest form the term independent of x in this expansion. ( also could you please check if this expansion of (a+b)^12 is correct?) a^12+12a^11+b+66a^10b^2+220a^9b^3........