HOW TO FIND THE CONSTANT TERM IN A BINOMIAL EXPANSION

Example 1 :

Find the constant term of (2x³ - (1/3x²))⁵

Solution :

  =  (2x³ - (1/3x²))⁵

 General term  T_r+1  =  ⁿC_r x^(n-r) a^r

x  = 2x³, n  =  5, a  =  (-1/3x²)

T_r+1  =  ⁵C_r (2x³)^5-r (-1/3x²)^r

  =  ⁵C_r (2)^5-r x^{15 - 3r} (-1/3)^r x^-2r

  =  ⁵C_r (-1/3)^r (2)^5-r x^{15 - 5r} 

Constant term : 

15 - 5r  =  0

15  =  5 r

r  =  15/5  =  3

= ⁵C₃ (-1/3)³ (2)^5-3 x^{15 - 5(3)} 

=  (-10/27) ⋅ 4  =  -40/27

So, the constant term is -40/27.

Example 2 :

Find the last two digits of the number 3⁶⁰⁰  

Solution :

3⁶⁰⁰   =  (3²) ³⁰⁰ 

  =  (9)³⁰⁰

  =  (10 - 1)³⁰⁰

  =  300^C₀ 10³⁰⁰ − 200^C₁ 10²⁹⁹ + · · ·+ 300^C₂₉₈ 10²(−1)²⁹⁸ + 300^C₂₉₉ 10(−1)²⁹⁹ + 300^C₃₀₀ (−1)³⁰⁰

= 10² [300^C₀ 10²⁹⁸ − 200^C₁ 10²⁹⁷ + · · ·+ 300^C₂₉₈ 10²(−1)²⁹⁸] - 3000 + 1

Multiple of 10 ends with 0. By subtracting 3000 from multiple of 10, we will get the value ends with 0.

Again by adding it by 1, we will get the value which ends with 01.

Example 3 :

If n is a positive integer, show that, 9ⁿ⁺¹ − 8n − 9 is always divisible by 64.

Solution :

(1 + x)ⁿ⁺¹  =  ⁽ⁿ⁺¹⁾C₀ + ⁽ⁿ⁺¹⁾C₀ x + ⁽ⁿ⁺¹⁾C₀ x² + ⁽ⁿ⁺¹⁾C₀ x³ + ^......+ ⁽ⁿ⁺¹⁾C_n+1 xⁿ⁺¹

(1+8)ⁿ⁺¹

  =  ⁽ⁿ⁺¹⁾C₀ + ⁽ⁿ⁺¹⁾C₁8 + ⁽ⁿ⁺¹⁾C₂(8)² + ⁽ⁿ⁺¹⁾C₃(8)³ + ^{...... ....}+  ⁽ⁿ⁺¹⁾C_n+1 8ⁿ⁺¹

9ⁿ⁺¹ − 8n − 9  =  [1 + (n+1)8 + 8² [⁽ⁿ⁺¹⁾C₂ + ⁽ⁿ⁺¹⁾C₃8+⁽ⁿ⁺¹⁾C₄8²

^{...... ....}+  ⁽ⁿ⁺¹⁾C_n+1 8^n-1] - 8n - 9

9ⁿ⁺¹ − 8n − 9  =  1 + 8n + 8 + [8² [⁽ⁿ⁺¹⁾C₂ + ⁽ⁿ⁺¹⁾C₃8+⁽ⁿ⁺¹⁾C₄8²

^{...... ....}+  ⁽ⁿ⁺¹⁾C_n+1 8^n-1] - 8n - 9

9ⁿ⁺¹ − 8n − 9  =  64 [An integer] 

9ⁿ⁺¹ − 8n − 9 is divisible by 64.

Example 4 :

If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)ⁿ are equal.

Solution :

If n is odd, then the two middle terms are T_(n−1)/2+1 and T_(n+1)/2+1

General term  T_r+1  =  ⁿC_r x^(n-r) a^r

x  =  x, n  =  n, a  =  y and r = (n−1)/2

ⁿC_{x  =}  T_(n−1)/2+1  =  ⁿC_(n−1)/2 x^(n-r) y^(n−1)/2  ----(1)

ⁿC_{y  =}  T_(n+1)/2+1  =  ⁿC_(n+1)/2 x^(n-r) y^(n+1)/2  ----(2)

If ⁿC_x  =  ⁿC_y  ==> then x + y = n

Evidently if x + y = n then ⁿC_x  =  ⁿC_y

  =  (n-1)/2 + (n + 1)/2

 =  2n/2

=  n

So, the coefficients of middle terms are equal.

Example 5 :

If n is a positive integer and r is a non negative integer, prove that the coefficients of x^r and x^n−r in the expansion of (1 + x)ⁿ are equal

Solution :

General term  T_r+1  =  ⁿC_r x^(n-r) a^r

x = 1, a = x, n = n 

By comparing the coefficients of (1) and (2)

ⁿC_r  =  ⁿC_n-r

So, they are equal.

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.