MCQ
The value of $(1+\text{i})^4+(1-\text{i})^4$ is:
- A8
- B4
- C-8
- D-4
Solution:
Using $\text{a}^4+\text{b}^4=(\text{a}^2+\text{b}^2)^2-2\text{a}^2\text{b}^2$
$(1+\text{i})^4+(1-\text{i})^4$
$=\Big((1+\text{i})^2+(1-\text{i})^2\Big)^2-2(1+\text{i})^2(1-\text{i})^2$
$=(1+\text{i}^2+2\text{i}+1+\text{i}^2-2\text{i})^2-2(1+\text{i}^2+2\text{i})(1+\text{i}^2-2\text{i})$
$=(1-1+2\text{i}+1-1-2\text{i})^2-2(1-1+2\text{i})(1-1-2\text{i})$
$=(0)-2(2\text{i})(-2\text{i}) \ (\because\text{i}^2=-1)$
$=8\text{i}^2$
$=-8$
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The two successive terms in the expansion of (1 + x)24 whose coefficients are in the ratio 1 : 4 are:
$[\text{Hint}:\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}\ \frac{\text{r}+1}{24-\text{r}}\ \frac{1}{4}\Rightarrow4\text{r}+4=24-4\Rightarrow\text{r}=4]$