What is the fourth term of a b8SolutionAs we know that ab2

Solution

As we know that (a+b)² = a²+2ab+b² and

(a+b)³ = (a+b)² * (a+b)

= (a²+2ab+b²)*(a+b)

= a³+3a²b+3ab²+b³ and this is called binomial expansion of (a+b)³ and if we multiply with (a+b) then we have the binomial expansion of (a+b)⁴. By this way we can derive a formula in a factorial from.

n! (n factorial) = if n is a positive integer, n! is defined to be product of all of the positive integer from 1 through n.

for example 3! = 3 * 2 * 1 so (a+b)⁴ are found by using factorilals:-

[4!/4!*0! *x² ] + [4!/(4-1)!*1! * x^4-1*y] + [4!/(4-2)!*2! * x^4-2y² ]+[4!/(4-3)!*3! * x^4-3y³ ] + [4!/0!*4! *y⁴ ]

So the binomial expansion of (a+b)ⁿ is given by:-

[n!/n!0! * xⁿ ] + [n!/(n-1)!1! * x^n-1y] + [n!/(n-2)!2! * x^n-2y²] + [.......] + [n!/0!n! * yⁿ]

and the binomial expansion of (a+b)⁸ is given by

[8!/8!0! * x⁸] + [8!/7!1! * x⁷y] + [8!/6!2! * x⁶y²] + [8!/5!3! * x⁵y³] +[.........] + [8!/0!8! * y⁸]

Hence the forth term is 8!/5!3! * x⁵y³