The nth term of the series 3 + 7 + 13 + 21 + 31 + ... is

Option c: n2 + n + 1. Solution: Let the nth term of the series be tn . The sequence of first differences is: 7-3=4, 13-7=6, 21-13=8, 31-21=10, ... Since the first differences are in A.P., the nth term is quadratic of the form tn = An2 + Bn + C . For n=1 : A + B + C = 3 For n=2 : 4A + 2B + C = 7 For n=3 : 9A + 3B + C = 13 Solving this system yields A = 1, B = 1, C = 1 . So the nth term is: tn = n2 + n + 1 Hence, Option C is the correct answer.

The $n^{th}$ term of the series $3 + 7 + 13 + 21 + 31 + \dots$ is

The correct answer is Option c: $n^2 + n + 1$. Let the $n^{\text{th}}$ term of the series be $t_n$. The sequence of first differences is: $$7-3=4, \ 13-7=6, \ 21-13=8, \ 31-21=10, \ \dots$$ Since the first differences are in A.P., the $n^{\text{th}}$ term is quadratic of the form $t_n = An^2 + Bn + C$. For $n=1$: $A + B + C = 3$ For $n=2$: $4A + 2B + C = 7$ For $n=3$: $9A + 3B + C = 13$ Solving this system yields $A = 1, B = 1, C = 1$. So the $n^{\text{th}}$ term is: $$t_n = n^2 + n + 1$$ Hence, **Option C** is the correct answer.

Sequence and SeriesMCQPYQ Jan 21Question 1872 of 212

All Questions

The $\displaystyle n^{th}$ term of the series $\displaystyle 3 + 7 + 13 + 21 + 31 + \dots$ is

Options

\displaystyle 4n - 1

\displaystyle n^2 + 2n

\displaystyle n^2 + n + 1

\displaystyle n^2 + 2

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Correct Answer

✅ Option c — $\displaystyle n^2 + n + 1$

All Options:

A $\displaystyle 4n - 1$
B $\displaystyle n^2 + 2n$
C $\displaystyle n^2 + n + 1$
D $\displaystyle n^2 + 2$

Detailed Solution & Explanation

Let the

\displaystyle n^{\text{th}}

term of the series be

\displaystyle t_n

.
The sequence of first differences is:

7-3=4, \ 13-7=6, \ 21-13=8, \ 31-21=10, \ \dots

Since the first differences are in A.P., the

\displaystyle n^{\text{th}}

term is quadratic of the form

\displaystyle t_n = An^2 + Bn + C

.
For

\displaystyle n=1

\displaystyle A + B + C = 3

For

\displaystyle n=2

\displaystyle 4A + 2B + C = 7

For

\displaystyle n=3

\displaystyle 9A + 3B + C = 13

Solving this system yields

\displaystyle A = 1, B = 1, C = 1

.
So the

\displaystyle n^{\text{th}}

term is:

t_n = n^2 + n + 1

Hence, **Option C** is the correct answer.

About This Chapter: Sequence and Series

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Arithmetic & Geometric Progressions

This chapter covers Arithmetic Progressions (AP) and Geometric Progressions (GP). Students learn how to find the nth term, sum of n terms, arithmetic/geometric means, and sum to infinity of a GP.

View Official ICAI Syllabus

Exam Strategy Tip

For complex 'sum of series' questions, a great hack is to substitute n = 1 and n = 2 into the question and the options to see which one matches, completely bypassing the formula.

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The nth\displaystyle n^{th}nth term of the series 3+7+13+21+31+…\displaystyle 3 + 7 + 13 + 21 + 31 + \dots3+7+13+21+31+… is