# Find the 24th term in the expansion of (a+b)^25?

Mar 27, 2018

${t}_{24} = - 300 \cdot {a}^{2} \cdot {b}^{23}$

#### Explanation:

We know that , in the expansion of ${\left(a + b\right)}^{n}$

${T}_{r + 1} = {\left(- 1\right)}^{r} {C}_{r}^{n} {\left(a\right)}^{n - r} {\left(b\right)}^{r}$

Comparing ${\left(a + b\right)}^{25} w i t h , {\left(a + b\right)}^{n}$

$n = 25 \mathmr{and} r + 1 = 24 \implies r = 23$

So,

${T}_{23 + 1} = {\left(- 1\right)}^{23} {C}_{23}^{25} {\left(a\right)}^{25 - 23} {\left(b\right)}^{23.} \ldots \to \left(I\right)$

Now ,${\left(- 1\right)}^{23} = - 1$

and ${C}_{r}^{n} = {C}_{n - r}^{n}$

$\implies {C}_{23}^{25} = {C}_{25 - 23}^{25} = {C}_{2}^{25} = \frac{25 \times 24}{2 \times 1} = 300$

Hence, from $\left(I\right)$

${t}_{24} = - 300 \cdot {a}^{2} \cdot {b}^{23}$