Question
If $\left(1+2 x+x^{2}\right)^{n}=\sum_{r=0}^{2 n} a_{r} x^{r}$, then $a_{r}=$(A) $\left({ }^{n} C_{r}\right)^{2}$(B) ${ }^{n} C_{r} \cdot{ }^{n} C_{r+1}$(C) ${ }^{2 n} C_{r}$(D) ${ }^{2 n} C_{r+1}$
Step 1
Step 1: We are given that $\left(1+2 x+x^{2}\right)^{n}=\sum_{r=0}^{2 n} a_{r} x^{r}$. Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 50 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
If $\sum_{r=1}^{n} r x^{r-1}=\frac{1}{(1-x)^{2}} \cdot\left\{1+a x^{n}+b x^{n+1}\right\}$, then (A) $a=(n+1)$ (B) $b=n$ (C) $a=-(n+1)$ (D) $b=-n$
For $n \in \mathbf{N}$, let $$ S(k)=\sum_{r=10}^{n} r^{k}\left({ }^{n} C_{r}\right)^{2} $$ then (a) $S(0)={ }^{2 \mathrm{M}} C_{n}$ (b) $S(1)=\frac{1}{2} n\left({ }^{2 n} C_{n}\right)$ (c) $S(2)=n^{2}\left[\frac{1}{2}\left({ }^{2 n} C_{n}\right)-{ }^{2 n-2} C_{n-2}\right]$ (d) $S(0)+S(1)+S(2)=2^{2 n-1}\left(2^{n} C_{n}\right)$
If $S_{n}=\sum_{r=0}^{n} \frac{1}{{ }^{n} C_{r}}$ and $t_{n}=\sum_{r=0}^{n} \frac{r}{n_{r}}$, then $\frac{t_{n}}{S_{n}}$ is equal to (A) $\frac{1}{2} n$ (B) $\frac{1}{2} n-1$ (C) $n-1$ (D) $\frac{2 n-1}{2}$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD