Find $n$th term and sum of $n$ terms of the series: $1+3+7+15+31+....$
- $n$th term $T_{n}=(2^{n}-1)$. Sum of $n$ terms$=2^{n}-1-n$
- $n$th term $T_{n}=(2^{n}-1)$. Sum of $n$ terms$=2(2^{n}-1)-n$
- $n$th term $T_{n}=(2^{n}-1)$. Sum of $n$ terms$=2(2^{n}-1)-2n$
- $n$th term $T_{n}=(2^{n}-1-n)$. Sum of $n$ terms$=2(2^{n}-1)-n-1$
Reveal answer
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B
Correct answer
Explanation
The series is 1, 3, 7, 15, 31... which is 2^1-1, 2^2-1, 2^3-1, 2^4-1, 2^5-1. Thus Tn = 2^n - 1. The sum is sum(2^k - 1) from k=1 to n = (2^(n+1) - 2) - n = 2(2^n - 1) - n.
AI explanation
The difference between consecutive terms yields the sequence 2, 4, 8, 16, which forms a geometric progression with a common ratio of 2. Because T_n = T_1 + (2^1 + 2^2 + ... + 2^(n-1)), we use the sum of a geometric progression to find T_n = 1 + 2(2^(n-1) - 1) = 2^n - 1. To find the sum of n terms, we sum this nth term formula: S_n = the sum of (2^k - 1) from k=1 to n, which evaluates to 2(2^n - 1) - n. The correct nth term is T_n = (2^n - 1) and the sum is 2(2^n - 1) - n.