Let $a_{i}$ be a sequence of numbers, is there an explicit formula for $\sum\limits_{i=0}^p \binom{p}{i}a_{i} $ involving only $p$ and $a_{n}$ ? Or is there any bound interms of $p$ and $a_{i}$ ?
Explicit formula involving Binomial coefficient
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binomial-coefficients
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1I don't believe such a formula exists in general, but a trivial upper bound is given by $$\sum_{i=0}^p \binom{p}{i} a_i \leq \sum_{i=0}^p \binom{p}{I} \sum_{i=0}^p a_i = 2^p \sum_{i=0}^p a_i$$ Or a slightly better one comes from Cauchy-Schwartz: $$\sum_{i=0}^p \binom{p}{i} a_i \leq \sqrt{ \sum_{i=0}^p \binom{p}{I}^2 \sum_{i=0}^p a_i^2} = \sqrt{ \binom{2p}{p} \sum_{i=0}^p a_i^2 } \leq \sqrt{ 4^p \sum_{i=0}^p a_i^2 } = 2^p \sqrt{\sum_{i=0}^p a_i^2}$$. – 2017-02-23
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It might not be exactly what you're looking for, but you can get some useful results from exponential generating functions. We will proceed formally, meaning we won't concern ourselves with issues of convergence.
Suppose we have two power series, $\sum_{n=0}^{\infty} a_n \frac{x^n}{n!}$ and $\sum_{n=0}^{\infty} b_n \frac{x^n}{n!}$ then their product may be expressed as $$ \sum_{n=0}^{\infty} c_n \frac{x^n}{n!}, \quad\text{where $c_n = \sum_{k=0}^{n} \binom{n}{k} a_k b_{n-k}$}$$ This isn't too difficult to prove if you want to give it a go.
Hence if we choose $b_n = 1$ for all $n$, then
$$e^x \sum_{n=0}^{\infty} a_n \frac{x^n}{n!} = \sum_{n=0}^{\infty} a_n \frac{x^n}{n!} \sum_{n=0}^{\infty} (1) \frac{x^n}{n!} = \sum_{n=0}^{\infty} d_n \frac{x^n}{n!}$$ where, as you wanted, $d_n = \sum_{k=0}^n \binom{n}{k} a_k$.
Now, we may use formal differentiation and setting $x=0$ to find coefficients $d_n$:
$$\left. \frac{d^n}{dx^n} \sum_{m=0}^{\infty} d_m \frac{x^m}{m!} \right|_{x=0} = \left. d_n + \sum_{m=k+1}^{\infty} d_m n! \frac{x^m}{m!}\right|_{x=0} = d_n = \left. \frac{d^n}{dx^n} \left(e^x \sum_{m=0}^{\infty} a_m \frac{x^m}{m!} \right)\right|_{x=0}$$
How do we use this in practice? Sometimes we might have nice closed forms for the expression $\sum_{n=0}^{\infty}a_n \frac{x^n}{n!}$, and in such circumstances we may be able to find closed form expression for the desired series. For example,
$\sum_{k=0}^{n} \binom{n}{k} = \sum_{k=0}^n \binom{n}{k}(1)$, so consider $a_n=1$ for all $n$. $$\sum_{n=0}^{\infty} a_n \frac{x^n}{n!} = \sum_{n=0}^{\infty} (1) \frac{x^n}{n!} = e^x$$ then $$ \sum_{k=0}^{n} \binom{n}{k} = \left. \frac{d^n}{dx^n} \left(e^x \sum_{m=0}^{\infty} a_m \frac{x^m}{m!} \right)\right|_{x=0} = \left. \frac{d^n}{dx^n} \left(e^{2x} \right)\right|_{x=0} $$ Now, each time we differentiate $e^{2x}$ we get the same thing with a multiple of $2$ out front, so $\frac{d^n}{dx^n} e^{2x} = 2^n e^{2x}$. Evaluating at $x=0$, $$d_n = \sum_{k=0}^n \binom{n}{k} = 2^n$$
Suppose $a_n = n!$, then $$\sum_{n=0}^{\infty} (n!)\frac{x^n}{n!} = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$$ then, $$ \sum_{k=0}^n \binom{n}{k} n! = \left.\frac{d^n}{dx^n}\left(\sum_{m=0}^{\infty} a_m \frac{x^m}{m!} e^x\right)\right|_{x=0} =\left.\frac{d^n}{dx^n}\frac{e^x}{1-x}\right|_{x=0}$$
This expression may have a closed form that you may be able to find through Taylor series, or other means, but this may give you an easy way to at least prove it through induction.