I keep messing up series whenever the sigma doesn’t start at 0. My brain loves off-by-one errors.
Example: I want S = sum_{k=1}^n 3^k. I know the geometric sum formula (r^{n+1}-1)/(r-1) when it starts at k=0. I lazily plugged r=3 and got (3^{n+1}-1)/2. For n=3, that gives (81-1)/2 = 40, but the actual sum 3+9+27 is 39. So I’m off by 1. I then “fixed” it by subtracting the k=0 term (which is 1): (3^{n+1}-1)/2 – 1 = (3^{n+1}-3)/2, which does give 39 for n=3. That feels like a hack, not a rule I can trust.
I also tried reindexing: let j = k-1, so sum_{k=1}^n 3^k turns into sum_{j=0}^{n-1} 3^{j+1}. But then I get tangled: do I pull out a factor of 3 or adjust the top to n or n-1? I keep second-guessing which exponent or bound to shift.
Same headache with arithmetic sums. For example, sum_{k=2}^5 2k should be 4+6+8+10=28. One attempt: count 4 terms and do 2*(1+2+3+4)=20 (wrong). Another attempt that worked: 2*((1+2+3+4+5) – 1) = 2*(15-1) = 28. It feels like I’m just guessing which chunk to subtract.
Is there a simple rule-of-thumb for reindexing and adjusting bounds so I stop being off by one? Like a quick checklist: how to shift the index, how to count terms, and when to subtract the missing start term(s). A plain, no-nonsense method I can do in my head would be ideal, with a tiny-number sanity check to avoid dumb mistakes.
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4 Responses
Here’s the plain rule I use: shift the start to 0, and carry along whatever factor the shift creates. For a geometric sum, let j = k − a so k = j + a and j runs 0 to n − a. Then r^k = r^{j+a} = r^a r^j, so sum_{k=a}^n r^k = r^a sum_{j=0}^{n−a} r^j. Now apply the 0-start formula to the inner sum: sum_{j=0}^{m} r^j = (r^{m+1} − 1)/(r − 1) with m = n − a. That gives sum_{k=a}^n r^k = r^a (r^{n−a+1} − 1)/(r − 1) = (r^{n+1} − r^a)/(r − 1). In my head I sometimes remember this as “there are n − a terms,” which is almost right for tracking the top exponent, even though the exact count is n − a + 1. Tiny check on your example: sum_{k=1}^n 3^k = (3^{n+1} − 3)/2, so for n = 3 it’s (81 − 3)/2 = 39. Subtracting the k = 0 term from the 0-start formula is the same rule in disguise: (3^{n+1} − 1)/2 − 1 = (3^{n+1} − 3)/2.
For arithmetic sums I do the same two moves. Either (a) count terms and use “number of terms times average of first and last,” or (b) reindex to start at 0 and use a known formula. Example: sum_{k=2}^5 2k. First term is 4, last is 10, and I count 5 − 2 = 3 “steps,” so there are 4 terms; average is (4 + 10)/2 = 7, so total is 4 × 7 = 28. Or reindex with j = k − 2: then k = j + 2, j = 0..3, and 2k = 2(j + 2) = 4 + 2j. Now sum_{j=0}^3 (4 + 2j) = 4·4 + 2(0 + 1 + 2 + 3) = 16 + 2·6 = 28. Quick mental checklist I use: shift to start at 0, pull out the factor created by the shift (r^a for geometric, or constants for arithmetic), set the new top index to n − a, and do a tiny-number sanity check (e.g., plug n = a and make sure a single term comes out).
Oh, I feel this one. Off-by-one gremlins love to hide in sigmas. My rule-of-thumb is: first count how many terms you actually have, then shift the index so it starts at 0, and only then use your favorite “starts-at-zero” formula. Concretely: if k runs from m to n, then the number of terms is N = n − m + 1. Let j = k − m, so j runs from 0 to N − 1 and k = j + m. Rewrite the summand in terms of j; very often you can factor out the part that depends on m. Also, “subtract the missing start term(s)” isn’t a hack-it’s exactly the same logic: sum over a bigger, easy range minus the sum of the excluded terms. A tiny-number sanity check I like: set N = 1 (so m = n). Your formula should collapse to just the single term f(m).
Geometric example: sum from k = m to n of r^k becomes r^m times sum from j = 0 to N − 1 of r^j. That gives the clean formula sum r^k = r^m (r^N − 1) / (r − 1), where N = n − m + 1. For your case m = 1, n = 3, r = 3: N = 3, so 3^1 (3^3 − 1) / (3 − 1) = 3 * (27 − 1) / 2 = 39. Your “subtract 1” fix is the same thing in disguise: sum k=1..n r^k = sum k=0..n r^k − r^0, which simplifies to r (r^n − 1) / (r − 1) and matches r^m (r^N − 1) / (r − 1) with m = 1, N = n. Quick check with N = 1: the formula gives r^m (r^1 − 1)/(r − 1) = r^m, as it should.
Arithmetic example: sum from k = m to n of (a k + b). Easiest head method: N = n − m + 1, first term = a m + b, last term = a n + b, so the sum is N times the average, i.e., N * (first + last) / 2. Equivalently, a * N * (m + n)/2 + b * N. For your sum k=2..5 of 2k: N = 4, first = 4, last = 10, so sum = 4 * (4 + 10)/2 = 28. Sanity check with N = 1 gives a m + b, which catches off-by-one slips. If I had to boil it down to one sticky note: 1) count terms N = n − m + 1, 2) reindex j = k − m so bounds are 0..N − 1, 3) factor any constants and apply the start-at-0 formula, 4) do the N = 1 check. I’m not sure there’s a simpler universal trick, but this one is reliable.
Quick checklist: for sum_{k=a}^b f(k), count terms N = b−a+1 and shift j = k−a so it’s sum_{j=0}^{N−1} f(a+j); then plug in your template-geometric r^k becomes r^a (r^N−1)/(r−1) (so sum_{k=1}^n 3^k = 3(3^n−1)/2 = (3^{n+1}−3)/2), and linear ck+d becomes c((a+b)N/2) + dN (e.g., sum_{k=2}^5 2k = 2((2+5)·4/2) = 28).
Analogy: it’s like sliding a ruler so its 0-mark sits at k=a-shift first, count the N tick marks, then read off the formula.
Rule-of-thumb: count the cookies-there are N = n − a + 1 terms from k = a to n; for geometric sums pull out r^a and reindex to 0..N−1 to get r^a (r^N − 1)/(r − 1), and for arithmetic sums use N × (first + last)/2 (then multiply any overall factor).
Example: sum_{k=1}^n 3^k = 3(3^n − 1)/2, so for n=3 it’s 3(27−1)/2 = 39; and sum_{k=2}^5 2k has N=4, first=4, last=10, so 4(4+10)/2 = 28.