MCQ
Let $a_{1}, a_{2}, a_{3}, \ldots$ be a G.P. such that $a_{1}<0$; $a_{1}+a_{2}=4$ and $a_{3}+a_{4}=16 .$ If $\sum\limits_{i=1}^{9} a_{i}=4 \lambda,$ then $\lambda$ is equal to
- ✓$-171$
- B$171$
- C$\frac{511}{3}$
- D$-513$
$\mathrm{r}^{2} \mathrm{a}_{1}+\mathrm{r}^{2} \mathrm{a}_{2}=16$
$\Rightarrow \mathrm{r}^{2}=4 \Rightarrow \mathrm{r}=-2 \quad$ as $\mathrm{a}_{1}<0$
and $a_{1}+a_{2}=4$
$a_{1}+a_{1}(-2)=4 \Rightarrow a_{1}=-4$
$4 \lambda=(-4)\left(\frac{(-2)^{9}-1}{-2-1}\right)=(-4) \times \frac{513}{3}$
$\Rightarrow \lambda=-171$
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$\operatorname{det}\left[\begin{array}{cc}\sum_{k=0}^n k & \sum_{k=0}^n{ }^n C_k k^2 \\ \sum_{k=0}^n{ }^n C_k k & \sum_{k=0}^n{ }^n C_k 3^k\end{array}\right]=0$, holds for some positive integer $n$. Then $\sum_{k=0}^n \frac{{ }^n C_k}{k+1}$ equals