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HETAV PATEL

Grade 11,

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help please...... explain fully......... reply soon

help please...... explain fully......... reply soon

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2 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
9 years ago
Hello Student,
In this question you have to find out

\sum_{r=0}^{n-1} 3^r -1

\sum_{r=0}^{n-1} 3^r - \sum_{r=0}^{n-1}1

=[3^0 + 3^1 + 3^2 + 3^3 + .... + 3^{n-1}] - [n]

=Now applu sum of n terms of a GP
[\frac{1(3^n -1)}{3-1}]- [n]
so answer is c
milind
23 Points
9 years ago
sir i have problem in sum second term that is 1 ….….….\sum_{r=0}^{n-1} 3^r - \sum_{r=0}^{n-1}1

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