(A) Given the $n^{th}$ term is $T_n = 2n - 1$. The sum of $n$ terms is given by $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (2k - 1)$. Using the summat

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(A) Given the $n^{th}$ term is $T_n = 2n - 1$. The sum of $n$ terms is given by $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (2k - 1)$. Using the summation formulas $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^{n} 1 = n$,we get: $S_n = 2 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1$ $S_n = 2 \left( \frac{n(n+1)}{2} \right) - n$ $S_n = n(n+1) - n$ $S_n = n^2 + n - n = n^2$.

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

Easy

If the $n^{th}$ term of a sequence is $T_n = 2n - 1$,then the sum of $n$ terms $S_n = \dots$

A
$n^2$
B
$\frac{n(n+1)}{2}$
C
$\frac{n(n+1)(2n+1)}{6}$
D
$n+2$

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Sequences and Series

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nth term of special series, Sum to n terms and Infinite number of terms

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Difficult

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Medium

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MediumAIEEE 2002

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If the $n^{th}$ term of a sequence is $T_n = 2n - 1$,then the sum of $n$ terms $S_n = \dots$

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