# Binomial Theorem for any index

BINOMIAL THEOREM FOR ANY INDEX:

$\large (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + ….+ \frac{n(n-1)…(n-r+1)}{r!}x^r + ….+ terms \; upto \; \infty$

Observations:

(i)Expansion is valid only when –1 < x < 1

(ii) nCr can not be used because it is defined only for natural number, so nCr will be written as $\large \frac{n(n-1)…(n-r+1)}{r!}$

(iii) As the series never terminates, the number of terms in the series is infinite.

(iv) General term of the series $\large (1+x)^{-n} = T_{r+1} = (-1)^r \frac{n(n+1)(n+2)….(n+r-1)}{r!}x^r$

(v) General term of the series $\large (1-x)^{-n} = T_{r+1} = \frac{n(n+1)(n+2)….(n+r-1)}{r!}x^r$

(vi) If first term is not 1, then make first term unity in the following way:

$\large (a+x)^n = a^n(1+\frac{x}{a})^n \; , if \; |\frac{x}{a}| < 1$

### Important Expansions:

(i) $\large (1+x)^{-1} = 1-x + x^2 – x^3 + ….+ (-1)^r x^r + ….$

(ii) $\large (1-x)^{-1} = 1+x + x^2 + x^3 + ….+ x^r + ….$

(iii) $\large (1+x)^{-2} = 1 – 2x + 3x^2 – 4x^3 + ….+ (-1)^r (r+1)x^r + ….$

(iv) $\large (1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + ….+ (r+1)x^r + ….$

(v) $\large (1+x)^{-3} = 1 – 3x + 6x^2 – 10x^3 + ….+ (-1)^r \frac{(r+1)(r+2)}{r!} x^r + ….$

(vi) $\large (1-x)^{-3} = 1 + 3x + 6x^2 + 10x^3 + ….+ \frac{(r+1)(r+2)}{r!} x^r + ….$

### MULTINOMIAL EXPANSION

In the expansion of (x1 + x2 + . . . + xn)m where m , n ∈ N and x1, x2 , . . ., xn are independent variables, we have

(i) Total number of terms in the expansion = m+n-1Cn-1

(iii) Sum of all the coefficient is obtained by putting all the variables xi equal to 1 and is nm

Illustration :Find the total number of terms in the expansion of (1 + a + b)10 and coefficient of a2b3.

Solution: Total number of terms = 10+3–1C3-1 = 12C2 = 66

Coefficient of a2b3 $\large = \frac{10!}{2! \times 3! \times 5!}$

= 2520

Illustration: Find the coefficient of a10b7c3 in the expansion of ( bc + ca + ab)10.

Solution: The general term in the expansion of ( bc + ca + ab)10 is

$\large \frac{10!}{r! s! t!} (bc)^r (ca)^s (ab)^t$

$\large = \frac{10!}{r! s! t!} a^{t+s} b^{r+t} c^{r+s}$

where r + s + t = 10

For the coefficient of a10b7c3 we set t + s = 10, r + t = 7, r + s = 3.

Since r + s + t = 10 we get r = 0, s = 3, t = 7.

Thus, the coefficient of a10b7c3 in the expansion of ( bc + ca + ab)10 is

$\large \frac{10!}{0! 3! 7!} = \frac{10 \times 9 \times 8}{3 \times 2 }$

= 120

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