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Summing series and the method of differences questions
Two ways to reduce a long sum to a short formula. Standard results for the sums of powers of r, and the method of differences, in which intermediate terms cancel in pairs and only terms from the beginning and the end remain.
6 original questions · 24 marks · the summing series and the method of differences notes · Further algebra and series
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Write down the standard results for the sum of the first n positive integers and the sum of their squares.
Worked answer
Σr = n(n + 1)/2 and Σr² = n(n + 1)(2n + 1)/6, both from r = 1 to n. B1 B1 for the two standard results. Everything polynomial in this topic reduces to these and the cubes formula.Using the standard results, show that Σ 4r³ from 1 to n equals n²(n + 1)², and check the formula at n = 3.
Worked answer
Σr³ = n²(n + 1)²/4, so multiplying by 4 clears the fraction: Σ4r³ = n²(n + 1)². At n = 3: 4(1 + 8 + 27) = 144 and 9 × 16 = 144. Agreed. M1 for quoting Σr³ = n²(n + 1)²/4, A1 for the printed result, B1 for the check at n = 3.Show that Σ r(r + 2) from 1 to n equals n(n + 1)(2n + 7)/6.
Worked answer
Split: Σr² + 2Σr = n(n + 1)(2n + 1)/6 + n(n + 1). Take out n(n + 1)/6: the bracket is (2n + 1) + 6 = 2n + 7, giving n(n + 1)(2n + 7)/6. Check at n = 3: 3 + 8 + 15 = 26, and 3 × 4 × 13/6 = 26. M1 for splitting into Σr² + 2Σr, A1 for both standard results, M1 for taking out n(n + 1)/6, A1 for the printed result.Given that 1/((2r − 1)(2r + 1)) = ½(1/(2r − 1) − 1/(2r + 1)), find Σ 1/((2r − 1)(2r + 1)) from 1 to n.
Worked answer
The telescope cancels every inner term: the sum is ½(1 − 1/(2n + 1)) = n/(2n + 1). Check at n = 3: 1/3 + 1/15 + 1/35 = 3/7, and n/(2n + 1) = 3/7. M1 for writing out enough terms, A1 for the surviving pair, A1 for ½(1 − 1/(2n + 1)), A1 for n/(2n + 1). Only the first positive and last negative fractions survive.Find Σ r² for r from n + 1 to 2n, simplifying your answer, and verify it at n = 2.
Worked answer
Subtract two full sums: Σ to 2n minus Σ to n = 2n(2n + 1)(4n + 1)/6 − n(n + 1)(2n + 1)/6 = n(2n + 1)(7n + 1)/6, after taking out n(2n + 1)/6. At n = 2: 9 + 16 = 25 and 2 × 5 × 15/6 = 25. M1 for the difference of two sums, A1 for both standard results, M1 for taking out n(2n + 1)/6, A1 for n(2n + 1)(7n + 1)/6. Sums that start above 1 are always a difference of two sums that start at 1.Express 1/(r(r + 2)) in partial fractions. Hence show that Σ 1/(r(r + 2)) from r = 1 to n is 3/4 − (2n + 3)/(2(n + 1)(n + 2)), and state the sum to infinity.
Worked answer
1/(r(r + 2)) = A/r + B/(r + 2) gives 1 = A(r + 2) + Br. Putting r = 0 gives A = ½ and r = −2 gives B = −½, so 1/(r(r + 2)) = ½(1/r − 1/(r + 2)).
Now write the terms out. Each negative part cancels the positive part two rows below, not one, so two terms survive at each end rather than one:
2Σ = (1/1 − 1/3) + (1/2 − 1/4) + (1/3 − 1/5) + … + (1/(n − 1) − 1/(n + 1)) + (1/n − 1/(n + 2)).
The survivors are 1 and 1/2 at the top and −1/(n + 1) and −1/(n + 2) at the bottom.
So Σ = ½[3/2 − 1/(n + 1) − 1/(n + 2)]. Combining the last two over a common denominator: 1/(n + 1) + 1/(n + 2) = (2n + 3)/((n + 1)(n + 2)).
Hence Σ = 3/4 − (2n + 3)/(2(n + 1)(n + 2)), as required. Check at n = 2: 3/4 − 7/24 = 11/24, and 1/3 + 1/8 = 11/24.
As n → ∞ the subtracted fraction tends to 0, so the sum to infinity is 3/4.
M1 for setting up the partial fractions, A1 for the printed pair, M1 for writing out enough terms, A1 for the four survivors, A1 for ½[3/2 − 1/(n + 1) − 1/(n + 2)], A1 for the printed result, B1 for the sum to infinity. Writing out enough terms to see both ends of the cancellation is the method mark. Assuming one survivor at each end, as in the gap-one case, gives 1/2 in place of 3/4 and loses everything after it.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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