Question
Let $f_{1}(n)=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$, then $f_{1}(1)+f_{1}(2)+f_{1}(3)$$+\ldots+f_{1}(n)$ is equal to(A) $n f_{1}(n)-1$(B) $(n+1) f_{1}(n)+n$(C) $(n+1) f_{1}(n)-n$(D) $n f_{1}(n)+n$
Step 1
We are asked to find the sum of $f_{1}(1)+f_{1}(2)+f_{1}(3)+\ldots+f_{1}(n)$. Show more…
Show all steps
Your feedback will help us improve your experience
Hast Aggarwal and 97 other Algebra 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
Let $(x+1)(x+2) \cdots(x+n)$ $=A_{0}+A_{1} x+A_{2} x^{2}+\cdots+A_{n} x^{n}$ then (a) $A_{0}-2 A_{1}+2^{2} A_{2}-\cdots+(-2)^{*} A_{n}=0$ (b) $A_{0}+2 A_{1}+3 A_{2}+\cdots+(n+1) A_{n}$ $\quad=(n+1) !\left(1+\frac{1}{2}+\cdots+\frac{1}{n+1}\right)$ (c) $n A_{0}+(n-1) A_{1}+\cdots+A_{n-1}$ $\quad=(n+1) !\left(\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}\right)$ (d) $A_{0}+A_{1}+\cdots+A_{n}=(n+1) !$
If $f(x+1)+f(x-1)=2 f(x)$ and $f(0)=0$ then find $f(n), n \in N .$
Real Function
Level III
$f f(x)=\frac{x-1}{x+1}$, then $f(2 x)$ in terms of $f(x)$ is (a) $\frac{\mathrm{f}(3 \mathrm{x})+1}{3 \mathrm{f}(\mathrm{x})-3}$ (b) $\frac{3 \mathrm{f}(\mathrm{x})+1}{\mathrm{f}(\mathrm{x})+3}$ (c) $\frac{2 \mathrm{f}(\mathrm{x})+1}{2 \mathrm{f}(\mathrm{x})-1}$ (d) $\frac{2 f(x)+1}{3 f(x)+2}$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD