Recover terms from a cumulative sum

The sum, SnS_n, of the first nn terms of a sequence U1,U2,U3,…U_1,U_2,U_3,\ldots is given by

Sn=n2(c−7n),S_n=\frac n2(c-7n),

where cc is a constant.

(i) Find UnU_n in terms of cc and nn. marks

(ii) Find a recurrence relation of the form Un+1=f(Un)U_{n+1}=f(U_n). marks

[2011/ACJC/Promo/13b]

Total marks: 4

Solution

(i)

Difference consecutive partial sums

Sn=n2(c−7n)S_n=\frac n2(c-7n) is the sum of U1,…,UnU_1,\ldots,U_n, where cc is a constant. Find UnU_n in terms of cc and nn.

Explore this idea

Hint

  1. Write UnU_n as Sn−Sn−1S_n-S_{n-1}.

  2. Replace nn by n−1n-1 in the formula for SnS_n before subtracting.

  3. Check n=1n=1, where S0=0S_0=0 is needed.

Answer · (i)

Un=(c+7−14n)/2U_n=(c+7-14n)/2.

Explore the idea

Each term is the step between consecutive partial sums

The running total Sₙ rises or falls by Uₙ at every step. Choose n to compare the step with the drop in the total.

Number of termsn = 3
Try an example
term Uₙ at n = 3
-2.5
sum Sₙ at n = 3
13.5
Key idea

Sn−Sn−1S_n-S_{n-1} is the vertical step of the sum curve at nn. That step is Un=c+7−14n2U_n=\frac{c+7-14n}{2}, here with c=30c=30.

Ideas to try
  1. At n=1n=1 the term and the sum coincide, matching U1=S1U_1=S_1.
  2. Move from n=2n=2 to n=3n=3. The total changes from 1616 to 13.513.5, a drop of exactly ∣U3∣=2.5|U_3|=2.5. Once UnU_n is negative, the total falls.
(ii)

Shift the general-term index

Find a recurrence relation of the form Un+1=f(Un)U_{n+1}=f(U_n).

Explore this idea

Hint

  1. Write Un+1U_{n+1} by replacing nn with n+1n+1.

  2. Compare it with UnU_n without choosing a particular nn.

Answer · (ii)

Un+1=Un−7U_{n+1}=U_n-7.

Explore the idea

The step between terms stays at −7

Uₙ₊₁ − Uₙ is the same at every n, which is the recurrence in graphical form.

Position nn = 4
Try an example
term Uₙ at n = 4
-9.5
step Uₙ₊₁ − Uₙ at n = 4
-7
Key idea

A constant step of −7-7 means Un+1=Un−7U_{n+1}=U_n-7 for every nn, whatever the value of cc.

Ideas to try
  1. Read UnU_n at two neighbouring values of nn and subtract. Does the answer ever leave −7-7?
  2. The terms fall along a straight line, so the first-difference line is flat.
Takeaway

Quadratic sums come from linear terms

A partial sum that is quadratic in nn with zero constant term has linear terms, so the sequence is an arithmetic progression. The coefficient −72-\tfrac72 of n2n^2 in SnS_n is half the common difference −7-7.