Sums and Products (Sommes et Produits)

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A complete set of vocabulary flashcards covering summation formulas, geometric series, index manipulation, double sums, factorials, and finite product rules.

20 Terms

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Sum of a Constant Sequence

For any constant a∈Ra \in \mathbb{R} and integers p≤np \le n, the sum of constant terms equals ∑k=pna=(n−p+1)a\sum_{k=p}^n a = (n - p + 1)a.

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Sum of the First nn Integers

The closed-form formula for the sum of the first nn positive integers: ∑k=1nk=n(n+1)2\sum_{k=1}^n k = \frac{n(n+1)}{2}.

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Sum of the Squares of the First nn Integers

The closed-form formula for the sum of the first nn squares: ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}.

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Sum of the Cubes of the First nn Integers

The closed-form formula for the sum of the first nn cubes: ∑k=1nk3=n2(n+1)24\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}.

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Geometric Sum from Index 0

For all n∈Nn \in \mathbb{N}, the geometric sum equals ∑k=0nqk=n+1\sum_{k=0}^n q^k = n+1 if q=1q = 1, and ∑k=0nqk=1−qn+11−q\sum_{k=0}^n q^k = \frac{1-q^{n+1}}{1-q} if q≠1q \neq 1.

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General Geometric Sum

For any q≠1q \neq 1 and n≥pn \ge p, the sum of geometric terms starting from index pp is ∑k=pnqk=qp1−qn−p+11−q\sum_{k=p}^n q^k = q^p \frac{1-q^{n-p+1}}{1-q}.

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Distributivity of Summation

For any constant λ∈R\lambda \in \mathbb{R} that does not depend on the index kk, ∑k=1nλxk=λ∑k=1nxk\sum_{k=1}^n \lambda x_k = \lambda \sum_{k=1}^n x_k.

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Associativity and Commutativity of Summation

The rule that allows splitting the sum of two combined terms into separate sums: ∑k=1n(xk+yk)=∑k=1nxk+∑k=1nyk\sum_{k=1}^n (x_k + y_k) = \sum_{k=1}^n x_k + \sum_{k=1}^n y_k.

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Index Shift by Translation (i=k+mi = k+m)

For natural numbers p≤np \le n and integer mm, translating the summation index by i=k+mi = k + m gives ∑k=pnxk+m=∑i=p+mn+mxi\sum_{k=p}^n x_{k+m} = \sum_{i=p+m}^{n+m} x_i.

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Index Shift by Reversal (i=m−ki = m-k)

For natural numbers p≤np \le n and integer m≥nm \ge n, reflecting the summation index by i=m−ki = m - k gives ∑k=pnxm−k=∑i=m−nm−pxi\sum_{k=p}^n x_{m-k} = \sum_{i=m-n}^{m-p} x_i.

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Additive Telescoping Sum

For any sequence of real numbers x0,…,xn+1x_0, \dots, x_{n+1}, consecutive terms cancel to give ∑k=0n(xk+1−xk)=xn+1−x0\sum_{k=0}^n (x_{k+1} - x_k) = x_{n+1} - x_0.

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Double Sum Interchange (Independent Indices)

When the ranges of indices are independent, the summation order can be swapped freely: ∑(i,j)∈[ ⁣[1,n] ⁣]×[ ⁣[1,m] ⁣]xi,j=∑i=1n∑j=1mxi,j=∑j=1m∑i=1nxi,j\sum_{(i,j) \in [\![1,n]\!] \times [\![1,m]\!]} x_{i,j} = \sum_{i=1}^n \sum_{j=1}^m x_{i,j} = \sum_{j=1}^m \sum_{i=1}^n x_{i,j}.

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Double Sum Interchange (1≤i≤j≤n1 \le i \le j \le n)

For dependent triangular indices with non-strict inequality: ∑1≤i≤j≤nxi,j=∑i=1n∑j=inxi,j=∑j=1n∑i=1jxi,j\sum_{1 \le i \le j \le n} x_{i,j} = \sum_{i=1}^n \sum_{j=i}^n x_{i,j} = \sum_{j=1}^n \sum_{i=1}^j x_{i,j}.

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Double Sum Interchange (1≤i<j≤n1 \le i < j \le n)

For dependent triangular indices with strict inequality: ∑1≤i<j≤nxi,j=∑i=1n−1∑j=i+1nxi,j=∑j=2n∑i=1j−1xi,j\sum_{1 \le i < j \le n} x_{i,j} = \sum_{i=1}^{n-1} \sum_{j=i+1}^n x_{i,j} = \sum_{j=2}^n \sum_{i=1}^{j-1} x_{i,j}.

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Link Between Single and Double Sums

For real numbers x1,…,xnx_1, \dots, x_n with n∈N∗n \in \mathbb{N}^*, the expanded square of a sum is given by (∑k=1nxk)2=∑k=1nxk2+2∑1≤i<j≤nxixj\left(\sum_{k=1}^n x_k\right)^2 = \sum_{k=1}^n x_k^2 + 2 \sum_{1 \le i < j \le n} x_i x_j.

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Factorial (n!n!)

For any natural integer nn, the integer defined by 0!=10! = 1 and for n∈N∗n \in \mathbb{N}^*, n!=∏k=1nk=1×2×⋯×nn! = \prod_{k=1}^n k = 1 \times 2 \times \dots \times n, which satisfies n!=n×(n−1)!n! = n \times (n-1)!.

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Product of a Constant Sequence

For all n∈N∗n \in \mathbb{N}^* and constant value λ\lambda, the product of constant terms is ∏k=1nλ=λn\prod_{k=1}^n \lambda = \lambda^n.

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Multiplicativity of the Product Operator

For all n∈N∗n \in \mathbb{N}^*, the product of term-by-term sequence products separates as ∏k=1n(xkyk)=(∏k=1nxk)(∏k=1nyk)\prod_{k=1}^n (x_k y_k) = \left(\prod_{k=1}^n x_k\right)\left(\prod_{k=1}^n y_k\right).

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Product of a Quotient

When no yky_k is zero, the product of quotients factors as ∏k=1nxkyk=∏k=1nxk∏k=1nyk\prod_{k=1}^n \frac{x_k}{y_k} = \frac{\prod_{k=1}^n x_k}{\prod_{k=1}^n y_k}.

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Multiplicative Telescoping Product

When no xkx_k is zero, for all n∈Nn \in \mathbb{N}, consecutive quotients cancel to yield ∏k=0nxk+1xk=xn+1x0\prod_{k=0}^n \frac{x_{k+1}}{x_k} = \frac{x_{n+1}}{x_0}.