Question
If $A={ }^{2 n} C_{0} \cdot{ }^{2 n} C_{1}+{ }^{2 n} C_{1}{ }^{2 \mathrm{n}-1} C_{1}+{ }^{2 n} C_{2}{ }^{2 \mathrm{n}-2} C_{1}+\ldots$, then $A$ is(A) 0(B) $2^{n}$(C) $n 2^{2 \mathrm{n}}$(D) 1
Step 1
Step 1: We are given the expression $A={ }^{2 n} C_{0} \cdot{ }^{2 n} C_{1}+{ }^{2 n} C_{1}{ }^{2 \mathrm{n}-1} C_{1}+{ }^{2 n} C_{2}{ }^{2 \mathrm{n}-2} C_{1}+\ldots$. Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 74 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 $A={ }^{2 \mathrm{n}} C_{0}{ }^{2 \mathrm{n}} C_{1}+{ }^{2 \mathrm{n}} C_{1}{ }^{2 \mathrm{n}-1} C_{1}+{ }^{2 \mathrm{n}} C_{2}{ }^{2 \mathrm{n}-2} \mathrm{C}_{1}+\ldots$, then $A$ is (A) 0 (B) $2^{\mathrm{n}}$ (C) $n 2^{2 \mathrm{n}}$ (D) 1
If $(1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{n} x^{n}$, then for $n$ even, $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-\ldots+(-1)^{n} C_{n}^{2}$ is equal to (A) 0 (B) $(-1)^{n / 2 n} C_{n / 2}$ (C) ${ }^{n} C_{n / 2}$ (D) none of these
If $\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}$, then value of $N=a_{0} p_{1}-a_{1} a_{2}+a_{2} o_{3}-a_{3} a_{4}+\cdots$ is equal to (a) 0 (b) $n$ (c) $-n$ (d) 1
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD