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Find the general term in the expansion of (1 – x)-3.

The r+1th term is given by \(\begin{array}{l}\mathbf{T}_{\mathrm{r}+1}=\frac{\mathbf{n}(\mathbf{n}-\mathbf{1})(\mathbf{n}-\mathbf{2}) \ldots(\mathbf{n}-\mathbf{r}+\mathbf{1})}{\mathbf{r} !} \mathbf{x}^{\mathbf{r}}\\ Now, \mathbf{T}_{\mathrm{r}+1}\\ =\frac{-3(-3-1)(-3-2)(-3-3) \ldots \ldots(-3-\mathbf{r}+1)}{\mathrm{r} !}(-\mathbf{x})^{\mathrm{r}}\\ =\frac{-3(-4)(-5)(-6) \ldots \ldots \cdot(-2-r)}{r... View Article