Coefficient of A Particular Term

In the expression of (a + x)ⁿ, the coefficient of second term is ⁿC₁, of the third term is ⁿC₂, of he fourth term is ⁿC₃ and so on. The suffix in each term being one less than the number of the term to which it applies; hence ⁿC_r is the coefficient of the (r+1)^th term. This is called the General Term, because by giving different numerical values to r any of the coefficients may be found. Also, the indices of a and x in the (r + 1)^th term are expressable in terms of r.

Thus, General term of the expansion (a + x)ⁿ is dented as

T_r+1 = ⁿC_r a^n-r x^r = ((n(n-1)(n-2)....(n-r+1))/r!)  a^n-r x^r.

Note: General term is useful in finding the following in Binomial expression.
(a) Particular term.
(b) Middle term
(c) Term independent of x.
(d) Term containing the greatest coefficient of x.

Illustration:

Find the fifth term in the expansion of (a + 2x³)¹⁷.

Solution:

General term of the expansion (a + x)ⁿ is

        T_r+1 = ⁿC_r a^n-r x^r.

Now, for the given expression, required term is

        T₅ = ¹⁷C₄ a¹³ (2x³)⁴.

        = 38080 a¹³ x¹².

Illustration:

Find the term independent of x in (1 + x + 2x³)  ((3/2)x² -  (1/3x))⁹.

Solution:

We want to find the term independent of x in product of (1 + x + 2x³) and  ((3/2)x² -  (1/3x))⁹ .  So, we should find the terms containing x⁰, x^-1 and x^-3 in the expansion  ((3/2)x² -  (1/3x))⁹  and multiply them respectively with the appropriate term from (1 + x + 2x³) i.e. constant term i.e. 1, term containing x i.e. 1 and the term containing x³ i.e. 2.

Now, T_r+1 = ⁿC_r a^n-r x^r [this is general term in (a+x)ⁿ]

                        = ⁹C_r ((3/2)x²)^9-r (-(1/3x))^r                      

This is general term in [((3/2)x² - (1/3x))⁹]

(i) To find the term containing x⁰ we use 18 - 3r = 0, i.e. r = 6

so coefficient of x⁰ we use (-1)^6 9C₆ 3^-3 2^-3 = ⁹C₆  (1/2².3³).

(ii) To find the term containing x^-1 we use 18-3r = -1 i.e. r = 19/3 (I) which is impossible.

(iii) To find the term containing x^-3 we put 18-3r = -3 i.e. r = 7. So, coefficient of x^-3 in ((3/2)x² - (1/3x))⁹ = (-1)^{7 9}C₇ 3^-5 2^-2 = -⁹C₇  (1/2³.3⁵) .

Therefore, required coefficient = ⁹C₆ (1/2³.3³)× 1-2 × ⁹C₇ (1/2³.3⁵) = 17/54.

 Illustration:

Find the coefficient of x²⁴ in  (x² +3a/x)¹⁵.

Solution:

The general term ((r + 1)th term) is

        (x² +3a/x)¹⁵ = ¹⁵C_r(x²)^15-r (3a/x)^r = ¹⁵C_r x^30-2r  (3^ra^r/x^r) = ¹⁵C_r 3^r a^r x^30-3r

If this term contains x²⁴, then 30-3r = 24

        => 3r = 6 => r = 2.

Therefore, the coefficient of x²⁴ = ¹⁵C₂ 9a².

Illustration:

If the binomial co-efficient of the (2r + 4)^th term and the (r - 2)^th term in the expansion of (1 + x)¹⁸ are equal, find the value r.

Solution:

The coefficient of (2r + 4)^th term in (1 + x)¹⁸ = ¹⁸C_2r+3 and the coefficient of (r-2)^th term = ¹⁸C_r-3, so that ¹⁸C_2r+3 = ¹⁸C_r-3.

        Either 2r + 3 = r - 3, => r = -6,

        Or 2r + 3 + r - 3 = 18 => 3r = 18 => r = 6.             (Since r ε N)

Illustration:

If the 4^th term in the expansion (px + 1/x)ⁿ is independent of x, find the value of n. Also calculate p if the 4^th term is 5/2.

Solution:

Here T₄ = T₃₊₁ = ⁿC₃(px)³ (1/x)^n-3 = ⁿC₃ p³ x^6-n.

T₄ is independent of x => 6 - n = 0 or n = 6.

Now given T₄ = 5/2 => ⁶C₃. p³ = 5/2 => p³ = 5/2 . = 1/8 => p = 1/2.

Illustration:

If the expansion of (1 + x)⁴³ the coefficient of (2r + 1)^th term is equal to the  coefficient of (r + 2)^th term find r.

Solution:

Given in the expansion of (1 + x)⁴³ the coefficient of (2r + 1)^th term = the coefficient of (r + 1)^th term.

          ⁴³C_2r = ⁴³C_r+1. Either 2r = r + 1 => r = 1,

        or 2r + r + 1 = 43 => r = 14.

        Hence r =1, 14

Illustration:

Find the coefficient of x³ in the expansion of (1 + x + 2x²) (2x² - (1/3x))⁹.

Solution:

        T_r+1 in  (2x² - (1/3x))⁹ is ⁹C_r (2x²)^9-r (-1/3x)^r = ⁹C_r2^9-r (-1/3)^r x^18-3r

        => coefficient of x³ is ⁹C₅.24 (-1/3)⁵ = -224/7.

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