Final answer:
The sum of the sequences 1, 5, 9, ..., (n+1) terms and 3, 7, 11, ..., n terms are found using the formula for the sum of an arithmetic series. Their respective sums are shown to have a ratio of (n+1):n when simplified.
Step-by-step explanation:
The student has asked to show that the sum of the sequence 1, 5, 9, ... to (n+1) terms has the ratio of (n+1):n to the sum of the sequence 3, 7, 11, ... to n terms. Let's denote the first sequence as S1 and the second sequence as S2.
For S1, the arithmetic sequence has a first term of 1 (a = 1) and a common difference of 4 (d = 4). The nth term of an arithmetic sequence is given by an = a + (n-1)d, so the nth term for S1 is 1 + (n-1)4 = 4n - 3. The sum of an arithmetic series can be found by S = (n/2) (2a + (n-1)d). Thus, the sum of S1 to (n+1) terms is S1 = ((n+1)/2) (2*1 + n*4) = ((n+1)/2) (2 + 4n) = (n+1)(n+2).
For S2, similarly, the first term is 3, and the common difference is also 4, the sum of S2 to n terms is S2 = (n/2) (2*3 + (n-1)*4) = (n/2) (6 + 4n - 4) = n(n+1).
Finally, to show the ratio of S1 to S2, we have S1/S2 = (n+1)(n+2) / n(n+1) = (n+2)/n = (n/n + 2/n) = 1 + 2/n, which simplifies to (n + 1):n as required.