Chanakya Niti In Oriya Pdfl ⏳

Chanakya Niti In Oriya Pdfl ⏳


Chanakya Niti In Oriya Pdfl

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Proving that $(n+1)! + (n+1) = (n+1)!$

I found this exercise on Catalan’s Numbers but I’m having a hard time on it.
Prove that $(n+1)! + (n+1) = (n+1)!$ where $n \in \mathbb{N}$.
I’ve tried the following:
Let $n = a + b$ where $a,b \in \mathbb{N}$.
$\displaystyle (a+1) (a+1)! + (a+1) = (a+1) (a+1)! + (a+1 + b)$.
This is exactly how it’s meant to be, but I don’t know how to prove it, because I don’t know how to proceed on this step.


$$\frac{(n+1)! + (n+1)}{n+1}= \frac{(n+1)!}{n+1}+1=\frac{(n+1)!}{n+1}+\frac{1}{n+1}=\ldots$$
using the obvious fact that


std::vector::data() and std::vector::operator[] for assignment

I have a question about how to use an std::vector for simple assignment.
The interface of the data() method seems to offer a way:
std::vector my_v = {1, 2, 3};

int i =;

but the vector is always accessed in the same manner as vector my_v = {1, 2, 3};.
Does this mean that copying the vector is done on the same basis as for:
int i = {1, 2, 3};

Does std::vector use the same way of checking for overload resolution for its data() and operator[]?


Yes, this means that both data() and operator[] are convertible to int (and that you

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