find the fourth term in the expansion of (x-4)^3

Thisis the Solution of Question From RD SHARMA book of CLASS 11 CHAPTER BINOMIAL THEOREM This Question is also available in R S AGGARWAL book of CLASS 11 Yo Thecoefficient of the x^2-term is 80. First, we need to find the sixth row of Pascal's Triangle to determine the coefficient of the non-simplified terms of the expansion of the binomial expression. Remember that Pascal's Triangle begins with 1 as the first row and as the first and last entry of every other row. The middle terms of each row are Transcript Question 3 Write the general term in the expansion of (x2 – y)6 We know that General term of expansion (a + b)n is Tr + 1 = nCr an–r br For (x2 – y)6 Putting n = 6 , a = x2 , b = –y Tr + 1 = 6Cr (x2)6 – r (–y)r = 6!/𝑟! (6 − 𝑟)! . (x)2 (6 – r) . (–1)r . (y)r = (–1)r 𝟔!/𝒓! (𝟔 − 𝒓)! . x12 – 2r Andthere you have it. I have just figured out the expansion of a plus b to the fourth power. It's exactly what I just wrote down. This term right over here, a to the fourth, that's what this term is. One a to the fourth b to the zero: that's just a to the fourth. This term right over here is equivalent to this term right over there. Thegeneral term in the given expansion is given by. T r+1 = (−1)r ×9Cr×(x3 2)9−r ×( 2 x2)r. Now pth term from the end. = (n−p+2)th term from the beginning. ∴ 4th term from the end. = (9−4+2)th term from the beginning. = 7th term from the beginning. = T 7 =T (6+1) = (−1)6×9C6×(x3 2)(9−6) ×( 2 x2)6 =9 C3×(x3 2)3 ×( 2 x2)6 10 Determine the new terms that would be added to \(P_3(x,y)\) (which you found in Exercise 13.7.1) to form \(P_4(x,y)\) and determine the fourth-degree Taylor polynomial for one of the functions we've considered and graph it together with the surface plot of the corresponding function in a 3D grapher like CalcPlot3D to verify that it paijaludea1975.

find the fourth term in the expansion of (x-4)^3