A half-tail identity and triangular rows

(a) The first 2 terms of a GP are p,qp,q, where q<p,p≠0q<p,p\ne0. If the sum of the first nn terms equals the sum to infinity of the remaining terms, prove pn=2qnp^n=2q^n. marks

(b) The AP 2,4,6,8,10,12,…2,4,6,8,10,12,\ldots is arranged in rows: 1st row: 2; 2nd: 4,6; 3rd: 8,10,12; 4th: 14,16,18,20; … (native table below).

(i) Show the first term in row nn is n2−n+2n^2-n+2. marks

(ii) Find the sum of all terms from row 1 to row n−1n-1. marks

[2009/DHS/Prelim/I/9]

Total marks: 8

Solution

(a)

Equal partial sum and tail

Use the equality of the first nn terms and the remaining infinite tail to prove pn=2qnp^n=2q^n.

Hint

  1. Use 2Sn=S∞2S_n=S_\infty, then cancel the common prefactor.

Answer · (a)

Equality of the partial sum and tail gives (q/p)n=1/2(q/p)^n=1/2, hence pn=2qnp^n=2q^n.

Nonzero convergent GP.

(b)(i)

First term in row n

Show that row nn begins with n2−n+2n^2-n+2.

Hint

  1. Count entries before the row; then add one for its first position.

Answer · (b)(i)

n2−n+2n^2-n+2.

(b)(ii)

Sum rows before n

Find the sum of all entries in rows 1 through n−1n-1.

Hint

  1. The last entry before row n is not the first entry of row n.

Answer · (b)(ii)

n(n−1)(n2−n+2)/4n(n-1)(n^2-n+2)/4.

Takeaway

Translate row positions into sequence indices

Although arranged as a table, all entries before row n remain the first N terms of 2,4,6,... .