MCQ
The value of $9^{\frac13}.9^{\frac19}.9^{\frac{1}{27}}\dots\text{to }\infty,$ is:
- A1
- B3
- C9
- DNone of these.
Solution:
$9^{\frac13}\times9^{\frac19}\times9^{\frac{1}{27}}\times\dots\infty$
$=9^{\big(\frac13+\frac19+\frac{1}{27}+\dots\infty\big)}$
Here, it is a G.P. with $\text{a}=\frac13$ and $\text{r}=\frac13.$
$\therefore9^{\Bigg(\frac{\frac{1}{3}}{1-\frac13}\Bigg)}$
$=9^{\big(\frac12\big)}=3$
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Let Sn denote the sum of the cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then $\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}$ equals to:
$\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
$\frac{\text{n}(\text{n}+1)}{2}$
$\frac{\text{n}^{2}+3\text{n}+2}{2}$
None of these.
The number of irrational terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is: