The binomial theorem or binomial expansion explains the algebraic expansion of the exponentiation of a binomial. It states that for any positive integer n, the sum of two numbers (say x & y) can be written as the sum of (n + 1) terms as,
(x + y)n = nC0xny0 + nC1xn-1y1 + nC2xn-2y2 + ….. nCn-1x1yn-1 + nCnxoyn
Binomial coefficients are positive integers that occur as coefficients in the binomial theorem. In a binomial expansion, there exist (n + 1) terms in total. The very first cases of the binomial theorem are as given below.
The number of terms in the (x + a)n ± (x – a)n is discussed below.
Let (x + a)n – (x – a)n —- (1)
Let (x + a)n + (x – a)n —- (2)
There can be 2 cases when the value of n is even and odd.
Case (1) (x + a)n – (x – a)n
In (x + a)n – (x – a)n, there are (n / 2) terms when the value of n is even.
In (x + a)n – (x – a)n, there are [(n + 1) / 2] terms when the value of n is odd.
Case (2) (1) (x + a)n + (x – a)n
In (x + a)n + (x – a)n, there are (n / 2) + 1 terms when the value of n is even.
In (x + a)n + (x – a)n, there are [(n + 1) / 2] terms when the value of n is odd.
Example: Find the number of terms in the expansion of (x + a)50 + (x – a)50.
Solution:
(a + x)n + (a – x)n = (a + b) (a – b) = 2a
= 2 [T1 + T3 + T5 ……. + T51]
= 2T1 + 2T3 + 2T5 + ……. + 2T51
= 26 terms
R-F Factor Relation
Aim: To find the integral and fractional part of the form (a + b √c)n, a, b, c belongs to the set of natural numbers.
Algorithm:
1) The given expression is of form I + f, where f is the fractional part that ranges from 0 to 1.
2) To find its conjugate f’, replace the positive sign with the negative sign.
3) Add or subtract to obtain an integral part = I
4) The remaining part becomes
When added,
0 < f < 1
0 < f’ < 1
Sum = 0 < f + f’ < 2
When subtracted,
0 < f < 1 and 0 < f’ < 1
– 1 < – f’ < 0
Difference = – 1 < f – f’ < 1
Example 1: The number of integral terms in the expansion of
Solution:
The general term is given by 256Cr . (√3)256−r . (8th root of √5)r
= 256Cr . (3)[256−r] / 2 . (5)r / 8
If [256 − r] / 2 is an integer, then (3)[256−r] / 2 is also an integer.
If r / 8 is an integer, then (5)r/8 is an integer.
The general term turns into an integer when [256 − r] / 2 and r / 8 both are integers.
[256 − r] / 2 becomes an integer when r can take values from 0, 2, 4, 6, ……, 256. [r / 8] becomes an integer when r can take values from 0, 8, 16, ……., 256.Combining the above two expressions, we can conclude that [256 − r] / 2 and r / 8 are integers when r = 0, 8, 16, …., 256 forms an AP.
∴256 = 0 + (n – 1) 8 (by the formula to find the nth term of an AP, tn = a + (n – 1) d)
n = 2568 + 1
= 32 + 1
= 33
Number of Terms and R-F Factor Relation – Video Lesson
Frequently Asked Questions
What is the total number of terms in a binomial expansion?
The total number of terms in a binomial expansion of (a + b)n is n + 1, i.e., one more than the exponent n. n ∈ N, a,b,∈ R.
What is the number of terms in the expansion (x + a)n + (x – a)n if n is even?
The number of terms in the expansion of (x + a)n + (x – a)n are (n+2)/2 if n is even.
What is the number of terms in the expansion (x + a)n + (x – a)n if n is odd?
The number of terms in the expansion of (x + a)n + (x – a)n are (n+1)/2 if n is odd.
What is the number of terms in the expansion (x + a)n – (x-a)n if n is even?
If n is even, the number of terms in the expansion of (x + a)n – (x-a)n is (n/2).
What is the number of terms in the expansion (x + a)n – (x-a)n if n is odd?
If n is odd, then the number of terms in the expansion of (x + a)n – (x-a)n is (n+1)/2.
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