Binomial Theorem

Binomial Expression

Any algebraic expression consisting of only two terms is known as a binomial expression.

It’s expansion in power of x is known as the binomial expansion.

e.g. (i) a + x (ii) a² + 1/x² (iii) 4x − 6y

Binomial Theorem
Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem.

For a positive integer n , the expansion is given by :

(a + x)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹ x + ⁿC₂ aⁿ⁻² x² + . . . + ⁿC_r a^n−r x^r + . . . + ⁿC_nxⁿ

where ⁿC₀, ⁿC₁ , ⁿC₂ , . . . , ⁿC_n are called Binomial co-efficients and there are (n + 1) terms in the expansion.

The value of ⁿC_r is defined as :

Similarly,

(a−x)ⁿ = ⁿC₀aⁿ − ⁿC₁aⁿ⁻¹ x + ⁿC₂ a^n-2 x² − . . . +( −1)^{r n}C_r a^n−r x^r + . . . + ( −1)^{n n}C_nxⁿ

Illustration : Expand

Solution:

⁷C₀x⁷ + ⁷C₁x⁶ + ⁷C₂x⁵ + ⁷C₃x⁴ + ⁷C₄x³ + ⁷C₅x² + ⁷C₆ x + ⁷C₇

= x⁷ + 7x⁵ + 21x³ + 35 x + 35/x + 21/x³ + 7/x⁵ + 1/x⁷

General Term in the Expansion:

The general term in the expansion of (a + x)ⁿ is the (r + 1)th term given as :

t_r+1 = ⁿCr a^n-r x^r .

Similarly the general term in the expansion of (x + a)ⁿ is given as :

t_r+1 = ⁿC_r x^n-r a^r .

The terms are considered from the beginning.

The (r + 1)th term from the end = (n − r + 1)th term from the beginning.

Corollary: Coefficient of x^r in expansion of (1 + x)ⁿ is ⁿC_r

Illustration : Find the co-efficient of x²⁴ in

Solution:

General term ((r + 1) th term) in

¹⁵C_r(x²)^15−r(3a/x)^r

= ¹⁵C_r x^30-2r(3^ra^r/x^r)

= ¹⁵C_r 3^ra^r x^30-3r

If this term contains x²⁴ .

Then 30−3r = 24

=> 3r = 6 => r = 2

Therefore, the co-efficient of x²⁴ = ¹⁵ C₂ . 9a²

Coefficients of Equidistant Terms from Beginning & end

The binomial coefficients in the expansion of (a + x)ⁿ equidistant from the beginning and the end are equal.
i.e. ⁿC_r = ⁿC_n−r

Note:

∎

∎ If ⁿC_x = ⁿC_y , then either x = y or x + y = n

Illustration 3: If the co-efficient of (2r + 4)th term and (r − 2)th term in the expansion of (1 + x)¹⁸are equal, find r

Solution: Since, co-efficient of (2r + 4)th term in (1 + x)¹⁸ = ¹⁸C_{2r + 3}

Co-efficient of (r − 2) th term = ¹⁸C_{r − 3}

=> ¹⁸C_{2r + 3} = ¹⁸C_{r − 3}

=> 2r + 3 + r − 3 = 18 => 3r = 18

=> r = 6

Illustration : Find the last three digits of 17²⁵⁶

Solution: We have 17² = 289 =290 − 1

Now, 17²⁵⁶ = (17²)¹²⁸ = (290 − 1)¹²⁸

= ¹²⁸C₀(290)¹²⁸ − ¹²⁸C₁(290)¹²⁷ + ……− ¹²⁸C₁₂₅(290)³ + ¹²⁸C₁₂₆(290)² − ¹²⁸C₁₂₇(290) + 1

, where m is a positive integer.

= 1000 m + (128) (290) [(127) (145) − 1 ] + 1 = 1000 m + (128) (290) (18414) + 1

= 1000 [m + 683527] + 680 + 1 = 1000 [m + 683527] + 681

Thus, the last three digits of 17²⁵⁶ are 681

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