Multiple choice

The sum of the series $4 + 8 + 16 + 32 + .......$. till $10$ terms is

  1. $6\, \times\, 2^{10}\, -\, 1$
  2. $4\, \times\, 2^{10}\, +\, 1$
  3. $4(2^{10}\, -\, 1)$
  4. $4(2^{10}\, +\, 1)$
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C Correct answer
Explanation

This is a geometric progression with a = 4, r = 2, and n = 10. The sum is a(r^n - 1) / (r - 1) = 4(2^10 - 1) / (2 - 1) = 4(2^10 - 1).

AI explanation

The series is a geometric progression with first term 4 and common ratio 2. Using the geometric progression sum formula, S = a times (r^n minus 1) divided by (r minus 1), we substitute the values to get S = 4 times (2^10 minus 1) divided by (2 minus 1). This simplifies directly to 4 times (2^10 minus 1).

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