-2
$\begingroup$

enter image description here

I've tried to prove at least one of them, but nothing is working. Could you give me a hint? Also i'd like to know something about properties of $\sum$, like multiplying them, etc(links).

$\endgroup$

closed as off-topic by John Doe, Matthew Conroy, Namaste, Rolf Hoyer, Krish Nov 26 '17 at 4:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Doe, Matthew Conroy, Namaste, Rolf Hoyer, Krish
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hint: For (7) and (8), consider the binomial expansion of $(x+1)^n$ and take derivatives. Then evaluate at $x = 1$. $\endgroup$ – eepperly16 Nov 25 '17 at 21:51
1
$\begingroup$

All of these identities can be derived from $$(1+x)^n = \sum_{k=0}^{n}\binom{n}{k}x^k. \tag{1}$$ For example, check how $(1)$ can be used to prove $5)$ here.

For $6)$, integrate $(1)$ as $$\int_{0}^1(1+x)^n dx = \frac{2^{n+1}-1}{n+1}=\sum_{k=0}^{n}\binom{n}{k}\int_{0}^{1}x^kdx=\sum_{k=0}^{n}\frac{\binom{n}{k}}{k+1}. \tag {2}$$

Likewise, take derivatives of $(1)$ w.r.t. $x$ to get $7)$ and $8)$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.