the first four terms in the expansion of (1-x)^n are 1-6x+ax^2+bx^3.find the n and also the respective value for a and b
(1-x)^n
= 1 + n(-x) + n(n-1)/2! (-x)^2 + n(n-1)(n-2)/3! (-x)^3 + ...
= 1-6x+ax^2+bx^3
so -nx = -6x
n = 6
n(n-1)/2! (-x)^2 = ax^2
6(5)/2=a
a = 15
n(n-1)(n-2)/3! (-x)^3 = bx^3
6(5)(4)/6 = -b
b = -20
To find the value of n, a, and b, we can apply the binomial expansion formula to the expression (1-x)^n.
The binomial expansion formula states that the terms in the expansion can be determined using the combination formula:
C(n, k) * a^(n-k) * b^k,
where C(n, k) represents the binomial coefficient, a represents the constant in the first term, and b represents the variable raised to a power.
In this case, we know the first four terms in the expansion of (1-x)^n are:
1 - 6x + ax^2 + bx^3.
We can identify the following information from these terms:
1st term: 1 (represents a^0, hence a = 1).
2nd term: -6x (represents a^1 * b^1, hence -6 = a * b).
3rd term: ax^2 (represents a^1 * b^2, hence a = a, b^2 = 0).
4th term: bx^3 (represents a^0 * b^3, hence b = b^3).
We can conclude the following:
n = 2 (Since the highest power of x in the expansion is x^2, n = 2).
a = 1 (From the first term).
b = -2 (By substituting a = 1 into the equation from the second term: -6 = a * b, -6 = 1 * b, b = -6).
Therefore, the value of n is 2, and the respective values for a and b are 1 and -2.