What is the coefficient of x ^99 in the expansion of (x-1) (x-2) (x-3) … (x-100)?
Let P(n) = (x-1).(x -2)….(x-n)So P(1) = x - 1P(2) = x^2 - (1 + 2).x + 2P(3) = x^3 - (1 + 2 + 3).x^2 + 11.x - 1.2.3So, P(n) = x^n - (1 + 2 + 3 +…+ n). x^(n-1) + … + ((-1)^n). 1.2.3…ncan be proved by method of induction.so, P(100) = x^100 - (1 + 2 + 3 +…+ 100). x^99 + … + 1.2.3…100So the coefficient of x^99 = -(1 + 2 + 3 + … + 100)as we know, the sum of first n natural numbers, 1 +2 +3 +… + n = (1 + n)*n/2So the coefficient of x^99 = - (1 + 100) * 100 / 2= - 101 * 50 = -5050
Find the coefficient of x^49 in the expansion of[math] (x-1)(x-3)(x-5)...(x-99).[/math]?
-A2A-I won't use any rocket science here. You can always figure out a pattern by considering fewer number of terms and then apply it as a generic pattern.[math](x-a)(x-b) \ = x^2 -(a+b)x + ab[/math][math](x-a)(x-b)(x-c) \ = x^3 -(a +b+c)x^2 +(ab + bc + ca)x - abc[/math]So, can you guess the pattern for product of 4 binomials?[math](x-a)(x-b)(x-c)(x-d) \ = x^4 - (a+b+c+d)x^3 + (ab +bc + cd +da)x^2 - (abc + bcd + cda + abd)x + abcd[/math]So, for the given product[math](x-1)(x-3)(x-5)\cdots \cdots(x-99) \ = x^{50} - (1+3+5+ \cdots + 99)x^{49} + (1*3 + 1*5 + 1*7 \cdots + 1*99 + 3*5 + 3*7 +\cdots 3*99 + 5*7 +\cdots + 97*99)x^{48} + \cdots[/math]Clearly, the answer is = - [Sum of odd numbers from 1 to 50] =[math] - \frac{50}{2}(1+99) \ = \large - 2500.[/math]I hope it helps!
The coefficient of x^2 in the binomial expansion of (1+x/2)^n, where n is a +ive integer, is 7?
So basically the value of n must satisfy this: nC2 * x^2/4 = 7x^2, we can remove the x^2's to solve the problem just in terms of n to get: nC2/4 = 7, so nC2 = 28. nC2 means n!/[(n-2)! 2!], putting this into the equation we can rearrange slightly to get: n!/(n-2)! = 56, n!/(n-2)! can be simplified to: n(n-1) so we can get a quadratic involving n to solve which is: n^2 -n - 56 = 0, which factorises to: (n - 8)(n + 7) = 0, so n = 8 or n = -7. As n must be a positive integer we know n = 8. The next part is simple to do, we know the value of n so the x^4 coefficient is: 8C4 * (x/2)^4. 8C4 = 70 and (x/2)^4 = x^4/16. So the coefficient of x^2 is: 70/16, or 35/8 if you prefer it simplified. Hopefully this helped.
What is the coefficient of [math]x^9[/math] in [math](1+x) (1+x^2) … (1+x^{100})[/math]?
Of the 100 terms, start multiplying first and last termsfor eg., (1+x)(1+x^100).. (1+x^2)(1+x^99) and so on..Now, we are left with 50 items..1+x+x100+x1011+x2+x99+x101..1+x50+x51+x101Ignore all the terms of power greater than 9 in each expression and start multiplying like we did before. ie., multiply 1st and 50th expressions.Repeat the above step again and reduce the number of terms to 25. Now again reduce to 13 items (12 reduced items and 13th will be left over item).Again reduce these 13 items to 7 items(6 reduced items and 13th is left over item)In each step, it is mandatory to exclude terms of power greater than 9.The final 7 items are 1+x, 1+x2, 1+x3, 1+x4, 1+x5+x9, 1+x6+x8, 1+x7 ..Now, Multiplying the above terms we will arrive at (1+2x^3+2x^4+2x^7+x+2x^5+x^8+x^2+2x^6+x^9)(1+x^7+x^5+x^6+x^8+x^9)So, the coefficient of x^9 is 7Please correct me if I missed something
What is the coefficient of [math]x^9[/math] in [math](1+x)(1+x^2)(1+x^3) …(1+ x^{100})?[/math]
First observe that [math]x^9[/math] can only be generated by products of the form [math]1.x^9, x^2.x^7 , x^4.x^3.x^2[/math] and so on.Therefore all we need to find is the number of ways the integer 9 can be partitioned into integers from the set[math]\{1,2,3,4,5,6,7,8,9\}[/math] . These are:[math]\{9\},\{1,8\},\{1,2,6\},\{1,3,5\},\{2,7\},\{3,6\},\{4,5\},\{2,3,4\}.[/math]There are 8 such terms.Therefore the coefficient of [math]x^9[/math] is 8.
The coefficient of x^3 is four times the coefficient of x^2 in the expansion of (1+x) ^n. Find the value of n. I appreciate any form of help.
A slightly complicated, yet interesting way to solve this question is by using Pascal’s Triangle.Pascal’s Triangle lists the coefficient of all the terms in a binomial expansion. For example, when n is 2, the binomial expansion of [math](x+1) [/math]is [math]x^2 + 2x + 1[/math], which corresponds to the third row of the triangle. Similarly, when n is 3, the binomial expansion of [math](x+1) [/math]is [math]x^3 + 3x^2 + 3x + 1[/math], which corresponds to the next row in the triangle.There are many patterns that can be found in a Pascal’s Triangle, such as the sum of two given numbers of one row will appear between the numbers on the next row. Also notice that no matter how much the triangle expands, the 4th number from the right side of the triangle will always be the coefficent for the term [math]x^3,[/math] and the number to the right of that will always be the coefficient for the term [math]x^2.[/math]One final observation that will help us answer this question is that we can see that when n increases by 1, the ratio between the coefficents for the terms [math]x^3[/math] and [math]x^2 [/math]always increases by [math]\frac{1}{3}.[/math] For example, on the fourth row of the triangle, when n is 3, the ratio between the coefficients is [math]1:3.[/math] For the next row (n = 4), the ratio for the coefficients of the two terms s [math]4:6[/math], or [math]2:3[/math]. And the next row after that (n =5 ), the ratio is [math]10:10[/math], or [math]3:3.[/math]Using this observation, all we have to do is find n when the ratio between the coefficents of the [math]x^2[/math] and [math]x^3[/math] is [math]12:3[/math], or 4. We can do so by using the following table:Therefore, the answer is n = 14.Hope that helps!
Binomial Expansions?
For all of these, you just need to know the binomial theorem (r+s)^n = Σ [k=0 to n] C(n,k) r^k s^(n-k) where C(n,k) is the binomial coefficient ("n choose k") given by the formula n!/(k!(n-k)!). The square brackets [k=0 to n] are just the limits of the summation. 1. This will be in the form of the theorem if r = 2p, s=(-3q), n=7. The term containing p^2 q^5 will be the term in the summation for k=2. So the entire term will be: C(7,2) (2p)^2 (-3q)^(7-2) = 7!/(2!5!) (2^2)(-3)^5 p^2 q^5 = 21*4*(-243) p^2 q^5 = -20412 p^2 q^5 2. Here, r = 2, s=-3q^2, n=4. The term containing p^4 = (p^2)^2 will be when n-k=2, i.e. k=2. So this term is: C(4,2) 2^2 (-3p^2)^2 = 4!/(2!2!)*4*9*p^4 = 108 p^4 3. Again, let s=3x, r=1/(6x), n=12. The constant term will appear whenever the power of r equals the power of s, so this means n-k = k, i.e. k=6. So the term is thus: C(12,6) 3^6 (1/6)^6 = 12!/(6!6!) 1/2^6 = (11*10*9*8*7)/128=433.125 4. r=1, s=-3x. The coefficient of x^2 will occur for the term where n-k = 2. So the term is: C(n,k) 1^(n-2) (-3x)^2 = 90 x^2 C(n,k) 9*x^2 = 90 x^2 C(n,k) = 10 n!/(k!(n-k)!) = 10 n!/(k!2!) = 10 n!/k! = 20 Don't forget n-k=2, so we also can write n!/(n-2)! = 20 n*(n-1) = 20 n^2-n = 20 n^2-n-20 = 0 (n-5)(n+4) = 0 n=5 or n-4 but n must be positive, so n=5 5. let r=ax, s=1, then the first three terms of the expansion are C(n,0) (ax)^0 1^n + C(n,1) (ax) 1^(n-1) + C(n,2) (ax)^2 1^(n-2) = 1 + n!/(n-1)! a x + n!/(2! (n-2)!) a^2 x^2 =1+n a x + n*(n-1)/2 a^2 x^2 Equating terms in the expansion we have n a = 24 and n(n-1)/2 a^2 =252 Solving we get n(n-1)/2 a^2 = (n a)^2/2 - n/2 a^2 = 24^2/2 - (24/a)/2 a^2 =288 - 12a = 252 a = 3 and n = 8 6. r = 2t^2, s=1/2, n=12. The middle term is where k=6: C(12,6) (2t^2)^6 (1/2)^6 = 12!/(6!6!) t^12 =1124 t^12
What is the binomial expansion of [math](1+x) ^{-2}[/math]?
Hope this will help to build fundamentals of binomial..
How do I calculate the coefficient of x to the power 9 in the polynomial [math]f(x) =(x-1) (x-2) (x-3) \ldots (x-10)?[/math]
At first think about this….How can you get x to the power 9 ???????in short multiplication like (x-1)(x-2)(x-3) how do you get x^2..yes, you are right , we get x^2 by multiplying two x and the integer of remaining terms. like, in the above example if u follow normal multiplication then you will multiply x of (x-1) and (x-2) and then -3 of (x-3) and you will get -3x^2.now in your problem if you want to get x^9 you have to multiply in this way . and you will get —x^9 — 2x^9 — 3x^9 —…..—10x^9so, the coefficient of x^9 is —1 — 2 — 3 — 4 ….—10= —55.Hope , you have understood. Thank you.