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Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) :

The sum of nn terms of the Progression

1+12+122+123+1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+ is 2n−1−12n−1\frac{2^{n-1}-1}{2^{n-1}}.

Reason (R) :

Sum of a geometric series having nn terms is given by Sn=a(1−rn)1−rS_{n}=\frac{a\left(1-r^{n}\right)}{1-r}, where aa is the 1st 1^{\text {st }} term and rr is the common ratio.

Solution

✅ Correct Option: 4
  1. First, let's verify the sum in Assertion (A):
  • We're given: 1+12+122+123+...1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + ... up to n terms
    • And it's claimed this equals 2n−1−12n−1\frac{2^{n-1}-1}{2^{n-1}}
  1. Reason (R) states:
  • For geometric series with n terms, Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r}
    • Where a is first term and r is common ratio
  1. Let's apply R's formula to the series in A:

a=1a = 1, r=12r = \frac{1}{2}

Sn=1(1−(12)n)1−12S_n = \frac{1(1-(\frac{1}{2})^n)}{1-\frac{1}{2}}

=1−12n12= \frac{1-\frac{1}{2^n}}{\frac{1}{2}}

=2(1−12n)= 2(1-\frac{1}{2^n})

=2−22n= 2-\frac{2}{2^n}

Which is NOT equal to 2n−1−12n−1\frac{2^{n-1}-1}{2^{n-1}}

  1. Therefore:
  • A is false
    • R is true (it's a standard formula)

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