Multiple choice

Find the sum of n terms of the series $3+8+22+72+266+1036+....$

  1. $\frac{3n}{4}+n(n+1)+\frac{1}{12}(4{n-1})$
  2. $\frac{3n}{4}+n(n+1)+\frac{1}{12}(4^{n-1})$
  3. $\frac{3n}{4}+n(n+1)+\frac{1}{12}(4{n-2})$
  4. $\frac{3n}{4}+n(n+1)+\frac{1}{12}(4^{n-2})$
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B Correct answer
AI explanation

Analyze the differences between consecutive terms by factoring them out relative to the previous term. The terms can be expressed using a combination of an arithmetic progression and powers of 4. Summing the general nth term of 4^(n-1) plus 2n plus 1 across n terms yields the total sum. Evaluating the geometric and arithmetic components leads to the sum formula 3n/4 + n(n+1) + (1/12)(4^n-1).

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