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Concept 01 / 07Sequences and Series

Sums of Natural Numbers, Squares and Cubes and Sigma Notation

Deriving sums of k, k^2 and k^3 by telescoping (k + 1)^m - k^m, properties of sigma notation, summing any polynomial general term, and finding the nth term of a series before summing.

Figure 1
Sigma notation writes a sum compactly: ∑k=1nTk=T1+T2+⋯+Tn\sum_{k=1}^{n}T_{k} = T_{1} + T_{2} + \cdots + T_{n}. It is linear: ∑(aTk+bUk)=a∑Tk+b∑Uk\sum(aT_{k} + bU_{k}) = a\sum T_{k} + b\sum U_{k}, and ∑k=1nc=nc\sum_{k=1}^{n}c = nc for a constant. But it does not distribute over products: ∑TkUk≠∑Tk∑Uk\sum T_{k}U_{k} \neq \sum T_{k}\sum U_{k} in general. The three standard sums. ∑k=1nk=n(n+1)2\sum_{k=1}^{n}k = \frac{n(n + 1)}{2}. ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n}k^{2} = \frac{n(n + 1)(2n + 1)}{6}. ∑k=1nk3=[n(n+1)2]2\sum_{k=1}^{n}k^{3} = \left[\frac{n(n + 1)}{2}\right]^{2}. The last one is remarkable: the sum of the first nn cubes is the square of the sum of the first nn natural numbers. Deriving them by telescoping. Sum the identity (k+1)2−k2=2k+1(k + 1)^{2} - k^{2} = 2k + 1 from k=1k = 1 to nn. The left side telescopes (every intermediate square cancels) to (n+1)2−1(n + 1)^{2} - 1. So 2∑k+n=n2+2n2\sum k + n = n^{2} + 2n, and ∑k=n(n+1)2\sum k = \frac{n(n + 1)}{2}. For squares, use (k+1)3−k3=3k2+3k+1(k + 1)^{3} - k^{3} = 3k^{2} + 3k + 1: summing gives (n+1)3−1=3∑k2+3n(n+1)2+n(n + 1)^{3} - 1 = 3\sum k^{2} + 3\frac{n(n + 1)}{2} + n, and solving yields the formula. For cubes, use (k+1)4−k4(k + 1)^{4} - k^{4}. This method derives the sum of any power from the sums of lower powers, and it is the same telescoping idea used for every 'method of differences' problem. Summing a polynomial general term. If TkT_{k} is a polynomial in kk, expand it, then use linearity and the three formulas. For example, ∑k=1nk(k+1)=∑k2+∑k=n(n+1)(2n+1)6+n(n+1)2=n(n+1)(n+2)3\sum_{k=1}^{n}k(k + 1) = \sum k^{2} + \sum k = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} = \frac{n(n + 1)(n + 2)}{3}. A shortcut for products of consecutive integers: ∑k=1nk(k+1)(k+2)⋯(k+r−1)=n(n+1)⋯(n+r)r+1\sum_{k=1}^{n}k(k + 1)(k + 2)\cdots(k + r - 1) = \frac{n(n + 1)\cdots(n + r)}{r + 1}. It follows by telescoping k(k+1)⋯(k+r)−(k−1)k⋯(k+r−1)=(r+1)k(k+1)⋯(k+r−1)k(k + 1)\cdots(k + r) - (k - 1)k\cdots(k + r - 1) = (r + 1)k(k + 1)\cdots(k + r - 1). Finding the general term first. To sum a series like 1⋅22+2⋅32+3⋅42+⋯1 \cdot 2^{2} + 2 \cdot 3^{2} + 3 \cdot 4^{2} + \cdots, identify the kkth term: here Tk=k(k+1)2T_{k} = k(k + 1)^{2}. Then expand and sum. Series of the form 1+(1+2)+(1+2+3)+⋯1 + (1 + 2) + (1 + 2 + 3) + \cdots have Tk=k(k+1)2T_{k} = \frac{k(k + 1)}{2}, a sum inside a sum; always simplify the inner sum first. Differences of consecutive terms. If the differences Tk+1−TkT_{k+1} - T_{k} form an AP, then TkT_{k} is quadratic in kk; if the second differences are constant, TkT_{k} is quadratic; if they form a GP, Tk=ark+bk+cT_{k} = a r^{k} + bk + c. Writing Tk=ak2+bk+cT_{k} = ak^{2} + bk + c and fitting three terms is often the quickest way to find a general term. Double sums. For ∑i=1n∑j=1n\sum_{i=1}^{n}\sum_{j=1}^{n}, the sum over all pairs factors if the summand is a product f(i)g(j)f(i)g(j). For restricted pairs, ∑i<jaiaj=(∑ai)2−∑ai22\sum_{i<j}a_{i}a_{j} = \frac{(\sum a_{i})^{2} - \sum a_{i}^{2}}{2}. For example, ∑1≤i<j≤nij=12[n2(n+1)24−n(n+1)(2n+1)6]\sum_{1 \le i < j \le n}ij = \frac{1}{2}\left[\frac{n^{2}(n + 1)^{2}}{4} - \frac{n(n + 1)(2n + 1)}{6}\right]. Key Takeaways: - ∑k=n(n+1)2\sum k = \frac{n(n + 1)}{2}; ∑k2=n(n+1)(2n+1)6\sum k^{2} = \frac{n(n + 1)(2n + 1)}{6}; ∑k3=[n(n+1)2]2\sum k^{3} = \left[\frac{n(n + 1)}{2}\right]^{2}. - Derivation: telescope (k+1)m+1−km+1(k + 1)^{m+1} - k^{m+1}. - ∑k=1nk(k+1)⋯(k+r−1)=n(n+1)⋯(n+r)r+1\sum_{k=1}^{n}k(k + 1)\cdots(k + r - 1) = \frac{n(n + 1)\cdots(n + r)}{r + 1}. - Find TkT_{k} first, expand, then sum term by term. - ∑i<jaiaj=(∑ai)2−∑ai22\sum_{i<j}a_{i}a_{j} = \frac{(\sum a_{i})^{2} - \sum a_{i}^{2}}{2}. - Sigma is linear but does not distribute over products. - Pitfall: summing nn terms when the series is written up to the nnth term of a different index. Brain-Teaser 1: Find the sum of nn terms of 1+(1+2)+(1+2+3)+⋯1 + (1 + 2) + (1 + 2 + 3) + \cdots. Solution: The kkth term is k(k+1)2\frac{k(k + 1)}{2}. So Sn=12∑k(k+1)=12⋅n(n+1)(n+2)3=n(n+1)(n+2)6S_{n} = \frac{1}{2}\sum k(k + 1) = \frac{1}{2} \cdot \frac{n(n + 1)(n + 2)}{3} = \frac{n(n + 1)(n + 2)}{6}. These are the tetrahedral numbers, 1,4,10,20,…1, 4, 10, 20, \ldots. Using the consecutive-product formula avoids expanding and recombining. Brain-Teaser 2: Find the sum of the series 12−22+32−42+⋯1^{2} - 2^{2} + 3^{2} - 4^{2} + \cdots to 2n2n terms. Solution: Group the terms in pairs: (12−22)+(32−42)+⋯=∑k=1n[(2k−1)2−(2k)2]=∑k=1n(1−4k)=n−2n(n+1)=−n(2n+1)(1^{2} - 2^{2}) + (3^{2} - 4^{2}) + \cdots = \sum_{k=1}^{n}[(2k - 1)^{2} - (2k)^{2}] = \sum_{k=1}^{n}(1 - 4k) = n - 2n(n + 1) = -n(2n + 1). Check with n=1n = 1: 1−4=−3=−1×31 - 4 = -3 = -1 \times 3. Pairing turns an alternating sum of squares into a linear sum, so no square-sum formula is needed at all. Brain-Teaser 3: Find the sum of nn terms of the series 3+7+13+21+31+⋯3 + 7 + 13 + 21 + 31 + \cdots. Solution: The differences are 4,6,8,104, 6, 8, 10, an AP, so TkT_{k} is quadratic. Write S=3+7+13+⋯+TnS = 3 + 7 + 13 + \cdots + T_{n} and also S=3+7+⋯+Tn−1+TnS = 3 + 7 + \cdots + T_{n-1} + T_{n} shifted by one place; subtracting gives 0=3+(4+6+⋯ to n−1 terms)−Tn0 = 3 + (4 + 6 + \cdots \text{ to } n - 1 \text{ terms}) - T_{n}. So Tn=3+(n−1)(n+2)=n2+n+1T_{n} = 3 + (n - 1)(n + 2) = n^{2} + n + 1. Then Sn=∑(k2+k+1)=n(n+1)(2n+1)6+n(n+1)2+n=n(n2+3n+5)3S_{n} = \sum(k^{2} + k + 1) = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} + n = \frac{n(n^{2} + 3n + 5)}{3}. Check at n=2n = 2: 2×153=10=3+7\frac{2 \times 15}{3} = 10 = 3 + 7.

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Sums of Natural Numbers, Squares and Cubes and Sigma Notation — DPP

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