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If nth term of a series is given by Tn = 3n + 2, where n is a natural number then find the value of \({S_n} = \;\mathop \sum \limits_{k = 1}^n {T_k} = \;?\)
1. \(\rm \frac{n(n+1)+4}{2}\)
2. \(\rm \frac{n(n+1)+2}{2}\)
3. \(\rm \frac{n(3n+7)}{2}\)
4. \(\rm \frac{3n(n+1)+4}{2}\)
5. None of these

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Correct Answer - Option 3 : \(\rm \frac{n(3n+7)}{2}\)

Concept:

  • Sum of the first n Natural Numbers \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)
  • Sum of the Square of the first n Natural Numbers \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
  • Sum of the Cubes of the first n Natural Numbers \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)
 

Calculation:

Given: nth term of a series is Tn = 3n + 2

Here, we have to find the value of \({S_n} = \;\mathop \sum \limits_{k = 1}^n {T_k} = \;?\)

Sum of series = \(\rm S_n = \sum (3n+2)=\sum 3n+\sum 2\)

As we know that, \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)

\(\Rightarrow \rm S_n =3\sum n+\sum 2\)

\(\Rightarrow \rm S_n =\frac{3n(n+1)}{2}+2n =\frac{n(3n+7)}{2}\)

Hence, option 3 is the correct answer

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