Math Buddyv1.0

Summation calculator with steps

Type or scan a sigma sum. MathBuddy splits it into known sums and adds it up step by step.

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What this page covers

  • Finite sums in sigma notation, starting at any index.
  • Sums of a constant, of kk, of k2k^2 and of k3k^3, using the standard closed formulas.
  • Geometric sums, finite and infinite, when the ratio allows an infinite total.
  • Not covered here: deciding whether a general series converges (use Series convergence calculator), though you can still type such a series and ask.

How to enter the problem

  • Type it with the sigma key on the math keyboard, or write it out, such as sum from k=1 to 10 of (2k+1).
  • Scan it. Fit the one problem in the frame.
  • Choose a photo. Several problems in one photo are solved as a list.
  • Then tap Solve.

Worked examples

Example 1

Problem

Evaluate ∑k=120k2\displaystyle \sum_{k=1}^{20} k^2.

Answer

Verified

The answer is

28702870

Explanation

  1. Recognise a sum of squares
    The first nn squares add up to n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6}.
  2. Substitute n=20n = 20
    20⋅21⋅416\frac{20 \cdot 21 \cdot 41}{6}.
  3. Simplify
    20⋅21=42020 \cdot 21 = 420 and 420÷6=70420 \div 6 = 70, so the sum is 70⋅41=287070 \cdot 41 = 2870.

Example 2

Problem

Evaluate ∑k=312(4k−5)\displaystyle \sum_{k=3}^{12} (4k - 5).

Answer

Verified

The answer is

250250

Explanation

  1. Count the terms
    From k=3k = 3 to k=12k = 12 there are 12−3+1=1012 - 3 + 1 = 10 terms.
  2. Split the sum
    ∑(4k−5)=4∑k−∑5\sum (4k - 5) = 4\sum k - \sum 5, both running from k=3k = 3 to 1212.
  3. Add the indices
    3+4+⋯+12=(3+12)⋅102=753 + 4 + \dots + 12 = \frac{(3 + 12) \cdot 10}{2} = 75
  4. Handle the constant
    The 5 is subtracted once per term
    5⋅10=505 \cdot 10 = 50
  5. Combine
    4⋅75−50=2504 \cdot 75 - 50 = 250

Example 3

Problem

Evaluate ∑k=0∞5(23)k\displaystyle \sum_{k=0}^{\infty} 5\left(\frac{2}{3}\right)^k.

Answer

Verified

The answer is

1515

Explanation

  1. Identify the series
    It is geometric. The first term (at k=0k = 0) is a=5a = 5 and the ratio is r=23r = \frac{2}{3}.
  2. Check the ratio
    ∣r∣<1|r| < 1, so the infinite sum has a finite value.
  3. Apply the formula
    a1−r=51/3=15\frac{a}{1 - r} = \frac{5}{1/3} = 15

Common mistakes

  • Counting terms as end minus start. From k=3k = 3 to k=12k = 12 there are 10 terms, not 9.
  • Adding a constant only once. In ∑k=110(2k+1)\sum_{k=1}^{10} (2k + 1) the +1+1 appears 10 times, so the total is 120, not 111.
  • Taking the first term of a geometric series at the wrong index. For ∑k=1∞5(23)k\sum_{k=1}^{\infty} 5\left(\frac{2}{3}\right)^k the first term is 103\frac{10}{3} and the sum is 10, not 15.
  • Using a1−r\frac{a}{1-r} when ∣r∣≥1|r| \ge 1. In that case the infinite sum has no finite value.

Checks, assumptions and limits

  • Check a small case by hand. ∑k=13(2k+1)=3+5+7=15\sum_{k=1}^{3}(2k+1) = 3 + 5 + 7 = 15, which matches the closed form n2+2nn^2 + 2n at n=3n = 3; at n=10n = 10 it gives 120.
  • The index letter does not matter: ∑k=110k\sum_{k=1}^{10} k and ∑i=110i\sum_{i=1}^{10} i are the same sum.
  • An infinite geometric sum exists only when ∣r∣<1|r| < 1.
  • MathBuddy can make mistakes. Double check important steps.

Related tasks

Frequently asked questions

Yes. Type the sum with the sigma key or scan it. The steps name the formula used for each part of the sum.

Any starting index works. The steps count the terms first, so a sum from 3 to 12 is treated as 10 terms.

For a geometric series with a ratio between −1 and 1, yes, using a/(1 − r). For other infinite series, ask, and the steps explain what can be said about the total.

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