Question

4. Find the square numbers that we will get if we add the following terms of triangular numbers series.​

Answer 1

Answer:

16

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Since all triangular numbers are of the form n(n+1)2, we must have the condition cited

m2=n(n+1)2(1)

Equation (1) is equivalent to

2=(2n+1)2−1(2m)2(2)

According to standard continued fraction theory, a rational approximation to 2–√ as good as (2) must also be an approximant for the continued fraction of 2–√. So we must find an overestimate for 2–√ with an even denominator (continued fraction approximants alternate between over- and under-estimates). The continued fraction for 2–√ is {1;2,2,2,…}, so the sequences of numerators and denominators for the approximants satisfy

ak=2ak−1+ak−2 and bk=2bk−1+bk−2(3)

where a0=a1=b1=1 and b0=0. Computing the first few approximants, we get

11,32,75,1712,4129,9970,…(4)

It follows from (3) that every other denominator is even, and that they correspond to the overestimates. Therefore, every other approximant yields a square triangular number.

Every other term of a sequence that satisfies (3), also satisfies

a2k=6a2k−2−a2k−4 and b2k=6b2k−2−b2k−4(5)

where a0=1, a2=3, b0=0, and b2=2. Let 2nk+1=a2k and 2mk=b2k. Applying (5) yields

nk=6nk−1−nk−2+2 and mk=6mk−1−mk−2(6)

where n0=m0=0 and n1=m1=1.

Using standard recurrence methods, we can solve (6) for mk to get

m2k=((3+22–√)k−(3−22–√)k42–√)2=(17+122–√)k+(17−122–√)k−232(7)

Thus, the sequence TSk=m2k.