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for $|x|<1$ : $$\frac{1}{1-x}=\sum_{n\geq 0} x^n$$ put $t=-x$ you get : $$f(t)=\frac{1}{1+t}=\sum_{n\geq 0} (-1)^n t^n=\sum_{n\geq 0} \frac{(-1)^nn!}{n!} t^n=\sum_{n\geq 0} \frac{f^{(n)}(0)}{n!} t^n$$ so $$f^{(n)}(0)=(-1)^n n!$$

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The result is correct if $\sum_{i=1}^na_i$ is a multiple of $n$. Let $a=\frac1n\sum_{i=1}^na_i$; we want to show that $$\sum_{i=1}^n\binom{a_i}r\ge n\binom{a}r\;.\tag{1}$$ Suppose that $a_i<a<a_k$. Then \begin{align*} \binom{a_i}r+\binom{a_k}r&=\binom{a_i}r+\binom{a_k-1}r+\binom{a_k-1}{r-1}\\ &\ge\binom{a_i}r+\binom{a_k-1}r+\binom{a_i}{r-1}... 1 I think you have your wires crossed; particularly it seems like you're thinking about how the sum of the first n odd numbers gives n^2. This is woefully incorrect. Here is how you should frame it: Suppose that x_1,\ldots, x_n\ge 2, then you want to show that x_1\cdots x_n is odd if and only if x_i is odd for all i. In terms of an induction ... 1 the result is correct but in the demonstration it is best not to confuse the symbol \varepsilon and \delta Since f is uniformly continuous then \forall\epsilon>0, \exists\delta>0 s.t. \forall x,x'\in X, d(x,x')<\delta\Rightarrow\rho(f(x),f(x'))<\epsilon But since (x_n) is Cauchy in X, then \forall\delta>0, \exists N s.t. \... 1 SupposeA=B+C$$If$$B^T=B, C^T=-C,$$then according to the known property of transposition of sum of matrices$$A^T=(B+C)^T=B^T+C^T=B+(-C)=B-C$$Now we have$$A=B+C \tag 1\\ A^T=B-C\tag 2\\$$Adding (1) to (2) gives$$B={(A+A^T)\over 2}\\ $$Subtracting (2) from (1) gives$$C={(A-A^T)\over 2}\\ 

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