The sequence 8, 24, 48, 80, 120........consists of the positive multiples of 8 each of which is one less than a perfect square, find the 2011th term. Divide it by 2012 and find the auotient.

Dear Student!

Here is the answer to your query.

The given sequence is

8, 24, 48, 80, 120 ...

and can be written as

3^{2}–1, 5^{2}–1, 7^{2}–1, 9^{2}–1, 11^{2}–1 ...

It can be observed that the n^{th }term of the given sequence is

(2*n* + 1)^{2} –1

∴ 2011^{th} term = (2 × 2011 + 1)^{2} – 1 = 16184528

2012^{th} term = (2 × 2012 + 1)^{2} – 1 = 16200624

and quotient when 2011^{th} term is divided by 2012^{th} term is zero as any term in the given sequence will always be greater than its previous term.

Cheers!

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