How can we find the coefficient of x^203 in the expansion of (x-1) (x^2-2) (x^3-3) … (x^20-20)?
Fist find the the maximum power of [math]x[/math] that is [math]\sum_{n=1}^{n=20}n = 210[/math]Now list all the possible combinations to get [math]203[/math], they are[math]\begin{align}203 &= 210 - (7)\\&=210 - (1+6)\\&=210 - (2+5)\\&=210 - (3+4)\\&=210 - (1+2+4)\end{align}[/math]Now find coefficients of [math]x^{203}[/math]In [math]7[/math][math](x^7 - 7)(x^{203} + ax^{202} +\dots) = - 7[/math]In [math]1,6[/math][math](x^1 - 1)(x^6 - 6)(x^{203} + ax^{202} +\dots) = 6[/math]In [math]2,5[/math][math](x^2 - 2)(x^5 - 5)(x^{203} + ax^{202} +\dots) = 10[/math]In [math]3,4[/math][math](x^3 - 3)(x^4 - 4)(x^{203} + ax^{202} +\dots) = 12[/math]In [math]1,2,4[/math][math](x^1 - 1)(x^2 - 2)(x^4 - 4)(x^{203} + ax^{202} +\dots) = - 8[/math]So the coefficient of [math]x^{203}[/math] is [math]-7 + 6 + 10 + 12 - 8 = 28 - 15 = 13[/math]
Find the coefficient of x^3 in the expansion of (1-3x)(1+2x)^6?
OK, so this can be a little tricky. There is one way to do this (the hard way) which is to simply multiply the whole thing out, which will find A LOT of extraneous information (i.e. you'll find the coefficients for EVERYTHING). The other way is to figure out what leads to x³ terms: First realize that you get the following: (1 + 2x)⁶ = ? + ?x + ?x² + ?x³ + ?x⁴ + ?x⁵ + ?x⁶ Second, realize that by multiplying the above by (1 - 3x) what you are really doing is this: (1 - 3x) * X --> distribute the X 1X - 3x * X --> X - 3x * X So we need to pick out the x³ terms from both of these: X --> find x³ term 3x * X --> find x³ term The first is easy, it's just the x³ term from binomial expansion: (1 + 2x)⁶ --> (6 c 3) * 1³ * (2x)³ --> (6 c 3) = 20 and (2x)³ = 2³x³ = 8x³ 20 * 8x³ = 160x³ So that's one part...but NOW, what about -3x * X? When we multiply the expanded binomial by -3x, WHAT TERM will generate an x³ term? That's easy, it's the x² term...because we would have: -3x * (... + ?x² + 160x³ + ...) --> -3x * ... + -3 * ? * x * x² + -3 * 160x * x⁴ + -3x * ... So, NOW we ALSO need the x² term (from the binomial expansion): (6 c 2) * 1⁴ * (2x)² = 15 * 4x² = 60x² --> multiply THAT by -3x -3x * 60x² = -180x³ BUT, we are going to add that to 1 * the x³ term from the ORIGINAL binomial expansion to get: 160x³ - 180x³ = -20x³ So the coefficient for the x³ term is: -20 Edit: (I think you mis-wrote some things, but it definitely seemed like you were on the right track)...but you missed a piece of it: Again, remember: (1 - 3x)(1 + 2x)⁶ = 1(1 + 2x)⁶ - 3x(1 + 2x)⁶ So...you need to find the x³ term from: (1 + 2x)⁶ --> (6 c 3) * 1³ * (2x)³ = 160x³ AND from: -3x(1 + 2x)⁶ --> means find the x² term from (1 + 2x)⁶ --> (6 c 2) 1⁴(2x)² = 60x² --> -3x * 60x² = -180x³ --> now add the two coefficients from the two different parts (to get -180 + 160 = -20)
How do you find the coefficient of x^3 in the expansion of (1+x+2x^2) (2x^2-1/3x)^9?
As per binomial theorem:Consider the expansion:In general any (k+1)th term in the expansion is given as :Thus in general the power of x is 18–3kFor the coefficient of x^3 substitute 18–3k=3 yields us k=5For k=5 we get the term:Thus only for the expansion the coefficient of x^3 is (-224/27)However the expansion is multiplied with [math]1+x+2x^2[/math]Following is the effect of multiplication on the powers of x (one by one):When the expansion is multiplied with 1, there is no effect on the powers of xWhen the entire expansion multiplied with x, all the powers of x in the expansion gets added to 1:ie:For [math]x^3[/math] we substitute 19–3k=3 which yields k=5.33 since ‘k’ can take only integer values we wont have [math]x^3[/math] term on multiplying the expansion x.Similarly on multiplying the expansion with [math]x^2[/math]In general power of x the becomes 20–3k. for [math]x^3[/math] term substituting 20–3k=3 we get k=5.667 which is again not possible.Thus we get [math]x^3 [/math]only when the expansion is multiplied with 1 and the multiplication [math]x^2[/math] and x wont give any [math]x^3[/math]term. Thus the coefficient of [math]x^3[/math] will remain the same i.e. −224/27
How do I find the coefficient of x^2 in the expansion of (1+x^2) (x/2-4/x) ^6 (binomial expansion)?
[math](1+x^2)(\dfrac{x}{2} - \dfrac{4}{x})^6 = T_1 * T_2[/math]Binomial expansion of [math]T_2 = (\dfrac{x}{2} - \dfrac{4}{x})^6 =[/math][math]\sum\limits_{r=0}^{6} \binom{6}{r} (-1)^r (\dfrac{x}{2})^r (\dfrac{4}{x})^{6-r}[/math][math]= \sum\limits_{r=0}^{6} \binom{6}{r} (-1)^r 2^{-r} 4^{6-r} x^{r-(6-r)} [/math][math]= \sum\limits_{r=0}^{6} \binom{6}{r} (-1)^r 2^{-r} 2^{2(6-r)} x^{r-(6-r)} [/math][math]= \sum\limits_{r=0}^{6} \binom{6}{r} (-1)^r 2^{-r+12–2r} x^{2r-6}[/math][math]= \sum\limits_{r=0}^{6} \binom{6}{r} (-1)^r 2^{12–3r} x^{2r-6}[/math]Interested in the coefficient of [math]x^2[/math]This can happen in [math]2[/math] ways:constant term of [math]T_1[/math] is multiplied by the coefficient of [math]x^2[/math] of [math]T_2[/math][math]=> 2r-6 = 2 => r = 4[/math]Constant of [math]T_1 = 1[/math]Coefficient of [math]x^2[/math] of [math]T_2[/math]:Substituting [math]r = 4[/math] in[math] \binom{6}{r} (-1)^r 2^{12–3r}[/math][math]\binom{6}{4} (-1)^4 2^{12–3(4)} = \dfrac{6*5}{1*2} * 1 * 2^0 = 15[/math]Coefficient of [math]x^2[/math] = Constant of [math]T_1[/math] * Coefficient of [math]x^2 [/math]of [math]T_2[/math] [math]= C_1 = 1*15 = 15[/math][math]x^2[/math] of [math]T_1[/math] is multiplied by constant term of [math]T_2[/math]Coefficient of [math]x^2[/math] of [math]T_1 = 1[/math]Constant term of [math]T_2[/math] = when [math]x[/math] is raised to the power of [math]0[/math][math]=> 2r-6 = 0 => r = 3[/math]Substituting [math]r = 3[/math] in [math]\binom{6}{r} (-1)^r 2^{12–3r}[/math][math]\binom{6}{3} (-1)^3 2^{12–3(3)} = \dfrac{6*5*4}{1*2*3} * (-1) * 2^3 = -160[/math]Coefficient of [math]x^2[/math] = Coefficient of [math]x^2[/math] of [math]T_1[/math] * Constant of [math]T_2 = C_2 = 1*(-160) = -160[/math]Coefficient of [math]x^2[/math] in the expanded form[math] = C_1 + C_2 = 15 - 160 = -145[/math]Ans: -145
Find the coefficient of x^3 y^8 in the expansion of (3x-2y^2)^7?
(3x-2y^2)^7 (3x)^7 + 7C1 (3x)^6 (-2y^2) + 7C2 (3x)^5 (-2y^2)^2 + 7C3 (3x)^4 (-2y^2)^3 + 7C4 (3x)^3 (-2y^2)^4 The expansion continues but we may sop here. The 5th term contains x^3 y^8 7C4 = 7C3 = 7 x 6 x 5 / 1 x 2 x 3 =35 3^3 = 27 (-2)^4 = 16 The term is (35) (27) (16) x^3 y^8 The coefficient of x^3 y^8 is (35)(27)(16) = 15,120
The coefficient of x in the expansion of ( x=1/ax^2)^7 is 7/3 find the possible values of a?
I think you mean (x-1/(ax^2))^7 general term is using binomial expansion the terms are (7CK) x^k(-1/(ax)^2)^(7-k) = (7Ck) x^(k)(-1/a)^(7-k) x^2(k-7) x^(3k-14) and as is it x term k = 5 so we get (7C5) (-1/a)^2 = 7/3 or 21 (1/a)^2 = (7/3) or (1/a)^2 = 1/9 or a= +/- 3
How can we find the coefficient of [math]x^3[/math] in the expansion of [math](1+x+x^2) ^n[/math]?
(1 + x + x^2)^n = ((1 - x^3))^n/((1 - x))^ncoefficient of x^r in (1 - x)^-n is (n+r-1)C(n-1)Hence the solution is coefficient of x^3 - coefficient of x^0 = (n+2)C(n-1) - n*(n-1)C(n-1) = (n+2)*(n+1)*n/6 - n
Find the coefficient of x^11y^3 term, in the expansion of (x+y)^14? (?
(x+y)^14 solution: k x^(n-m) * y^m n = 14, m = 3 [ n! / (n-m)!*m! ] * (x^11)(y^3) [ 14! / (11!)(3!) ] x^11*y^3 = (14*13*12*11!)/(11!)(3!) [ x^11*y^3] = (14*13*12)/(3*2*1) [x^11*y^3] = (2184/6)x^11*y^3 = 364 x^11 y^3 hence, coefficient is 364