# Binomial Theorem

## Binomial Theorem

Binomial theorem: If |x| < 1 and p, q are positive integers then

$${{(1+x)}^{p/q}}=1+\frac{p}{1!}\left( \frac{x}{q} \right)+\frac{p(p-q)}{1.2}{{\left( \frac{x}{q} \right)}^{2}}+…+\frac{p(p-q)…(p-(r-1)q)}{1.2….r}{{\left( \frac{x}{q} \right)}^{r}}$$.

$${{(1-x)}^{p/q}}=1-\frac{p}{1!}\left( \frac{x}{q} \right)+\frac{p(p-q)}{1.2}{{\left( \frac{x}{q} \right)}^{2}}-…+{{(-1)}^{r}}\frac{p(p-q)…(p-(r-1)q)}{1.2….r}{{\left( \frac{x}{q} \right)}^{r}}$$.

$${{(1-x)}^{-p/q}}=1+\frac{p}{1!}\left( \frac{x}{q} \right)+\frac{p(p+q)}{1.2}{{\left( \frac{x}{q} \right)}^{2}}+…+\frac{p(p+q)…(p+(r-1)q)}{1.2….r}{{\left( \frac{x}{q} \right)}^{r}}$$.

$${{(1+x)}^{-p/q}}=1-\frac{p}{1!}\left( \frac{x}{q} \right)+\frac{p(p+q)}{1.2}{{\left( \frac{x}{q} \right)}^{2}}-…+{{(-1)}^{r}}\frac{p(p+q)…(p+(r-1)q)}{1.2….r}{{\left( \frac{x}{q} \right)}^{r}}$$.

Obtain the values of x for which the binomial expansions for the follows

Examples 1: (6 + x)3/2

Solution: Given that (6 + x)3/2

= $${{\left( 6 \right)}^{3/2}}{{\left( 1+\frac{x}{6} \right)}^{3/2}}$$,

Hence, the binomial expansion for (6 + x)3/2 is valid when |x/6|<1 ⇒ |x| < 6.

⇒ – 6 < x < 6

⇒ x ϵ (-6, 6)

Examples 2: (2 + 3x)⁻

Solution: Given that (2 + 3x)⁻

= $${{\left( 2 \right)}^{-2/3}}{{\left( 1+\frac{3x}{2} \right)}^{-2/3}}$$,

Hence, the binomial expansion for (2 + 3x)⁻ is valid when |3x/2| < 1.

⇒ |x| < ⅔

⇒ – ⅔ < x < ⅔

⇒ x ϵ (-⅔, ⅔).