AKUEB Class 11 (HSSC-I) Mathematics Miscellaneous Series MCQs & Practice Questions | AKUEB Prep

Miscellaneous Series - Class 11 (HSSC-I) Mathematics

7 practice questions for this chapter, gradeable instantly.

ClassSubjectChapterTopicSLODifficultyTypePast paperSort

Class 11 - Mathematics / Unit 4 - Miscellaneous Series / 4.1 Evaluation of $\sum n$, $\sum n^2$, $\sum n^3$ / 4.1.2

Past Paper 2024MEDIUM

The sum of the cubes of the first $10$ natural numbers is

55

385

2{,}025

3{,}025

Class 11 - Mathematics / Unit 4 - Miscellaneous Series / 4.1 Evaluation of $\sum n$, $\sum n^2$, $\sum n^3$ / 4.1.3

Past Paper 2024MEDIUM

$\displaystyle\sum_{n=1}^{10} n - \sum_{n=1}^{5} n$ is equal to

0

2

40

55

Class 11 - Mathematics / Unit 4 - Miscellaneous Series / 4.1 Evaluation of $\sum n$, $\sum n^2$, $\sum n^3$ / 4.1.3

Past Paper 2025MEDIUM

Which of the following options is equivalent to $\displaystyle\sum_{1}^{n}\left(2n^2 - n + 2\right)$ ?

2\displaystyle\sum_{1}^{n} n^2 - \sum_{1}^{n} n + 2

2\displaystyle\sum_{1}^{n} n^3 - \sum_{1}^{n} n^2 + 2

2\displaystyle\sum_{1}^{n} n^2 - \sum_{1}^{n} n + 2n

2\displaystyle\sum_{1}^{n} n^3 - \sum_{1}^{n} n^2 + 2n

Class 11 - Mathematics / Unit 4 - Miscellaneous Series / 4.1 Evaluation of $\sum n$, $\sum n^2$, $\sum n^3$ / 4.1.3

Past Paper 2025MEDIUM

If $\displaystyle\sum_{k=1}^{2} k = 3$ , then the value of $\displaystyle\sum_{k=3}^{n} k$ is equal to

\dfrac{n(n+1)}{2} - 3

n\left(n+\dfrac{1}{2}\right) - 3

3 - \dfrac{n(n+1)}{2}

3 - n\left(n+\dfrac{1}{2}\right)

Class 11 - Mathematics / Unit 4 - Miscellaneous Series / 4.1 Evaluation of $\sum n$, $\sum n^2$, $\sum n^3$ / 4.1.2

Model Paper 2026EASY

$\displaystyle\sum_{n=1}^{3}n^3$ is equal to

6

12

27

36

Class 11 - Mathematics / Unit 4 - Miscellaneous Series / 4.1 Evaluation of $\sum n$, $\sum n^2$, $\sum n^3$ / 4.1.3

Model Paper 2026MEDIUM

If $\displaystyle\sum_{k=1}^{2}k=3$ , then the value of $\displaystyle\sum_{k=3}^{n}k$ is equal to

\dfrac{n(n+1)}{2}-3

n\left(n+\dfrac{1}{2}\right)-3

3-\dfrac{n(n+1)}{2}

3-n\left(n+\dfrac{1}{2}\right)

Class 11 - Mathematics / Unit 4 - Miscellaneous Series / 4.1 Evaluation of $\sum n$, $\sum n^2$, $\sum n^3$ / 4.1.3

Model Paper 2026EASY

$\displaystyle\sum_{n=1}^{10}n-\sum_{n=1}^{5}n$ is equal to

15

40

55

70