Summation of Special Series — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Summation of Special Series MCQs with step-by-step solutions (20 questions). Part of Sequences and Series. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Summation of Special Series · easy · numerical
The sum of the first $100$ natural numbers $\sum_{r=1}^{100} r$ is:
A. $10100$
B. $5050$ ✓ Correct
C. $4950$
D. $5000$
Solution: $\dfrac{n(n+1)}{2} = \dfrac{100 \cdot 101}{2} = 5050$.
Q2 — Summation of Special Series · medium · numerical
The value of $\sum_{r=1}^{10} r^2$ is:
A. $385$ ✓ Correct
B. $285$
C. $400$
D. $505$
Solution: $\dfrac{n(n+1)(2n+1)}{6} = \dfrac{10 \cdot 11 \cdot 21}{6} = 385$.
Q3 — Summation of Special Series · medium · numerical
The value of $\sum_{r=1}^{5} r^3$ is:
A. $100$
B. $125$
C. $255$
D. $225$ ✓ Correct
Solution: $\left[\dfrac{n(n+1)}{2}\right]^2 = \left(\dfrac{5 \cdot 6}{2}\right)^2 = 15^2 = 225$.
Q4 — Summation of Special Series · medium · numerical
The sum $1 + 3 + 5 + \cdots + (2n-1)$ for $n = 15$ is:
A. $240$
B. $196$
C. $210$
D. $225$ ✓ Correct
Solution: The sum of the first $n$ odd numbers is $n^2 = 225$.
Q5 — Summation of Special Series · hard · numerical
The value of $\sum_{r=1}^{n} r(r+1)$ for $n = 5$ is:
A. $55$
B. $70$ ✓ Correct
C. $60$
D. $85$
Solution: $\sum r(r+1) = \dfrac{n(n+1)(n+2)}{3} = \dfrac{5 \cdot 6 \cdot 7}{3} = 70$.
Q6 — Summation of Special Series · hard · numerical
The value of $\sum_{r=1}^{3} r(r+1)(r+2)$ is:
A. $100$
B. $60$
C. $120$
D. $90$ ✓ Correct
Solution: $\dfrac{n(n+1)(n+2)(n+3)}{4} = \dfrac{3 \cdot 4 \cdot 5 \cdot 6}{4} = 90$ (check: $6 + 24 + 60 = 90$).
Q7 — Summation of Special Series · medium · numerical
The sum of the squares of the first $5$ even numbers $2^2 + 4^2 + \cdots + 10^2$ is:
A. $220$ ✓ Correct
B. $180$
C. $200$
D. $240$
Solution: $= 4(1^2 + 2^2 + \cdots + 5^2) = 4 \cdot 55 = 220$.
Q8 — Summation of Special Series · hard · numerical
The sum $1^2 + 3^2 + 5^2$ (first three odd squares) is:
A. $35$ ✓ Correct
B. $30$
C. $45$
D. $25$
Solution: $1 + 9 + 25 = 35$ (matching $\dfrac{n(2n-1)(2n+1)}{3} = \dfrac{3 \cdot 5 \cdot 7}{3}$).
Q9 — Summation of Special Series · hard · numerical
If $1^2 + 2^2 + \cdots + n^2 = 285$, then $n$ equals:
A. $7$
B. $10$
C. $9$ ✓ Correct
D. $8$
Solution: $\dfrac{n(n+1)(2n+1)}{6} = 285 \Rightarrow n(n+1)(2n+1) = 1710 = 9 \cdot 10 \cdot 19 \Rightarrow n = 9$.
Q10 — Summation of Special Series · hard · numerical
If $1^3 + 2^3 + \cdots + n^3 = 3025$, then $n$ equals:
A. $9$
B. $12$
C. $11$
D. $10$ ✓ Correct
Solution: $\left[\dfrac{n(n+1)}{2}\right]^2 = 3025 \Rightarrow \dfrac{n(n+1)}{2} = 55 \Rightarrow n(n+1) = 110 \Rightarrow n = 10$.
Q11 — Summation of Special Series · medium · numerical
If $\sum_{r=1}^{n} r = 78$, then $n$ equals:
A. $12$ ✓ Correct
B. $13$
C. $11$
D. $14$
Solution: $\dfrac{n(n+1)}{2} = 78 \Rightarrow n(n+1) = 156 = 12 \cdot 13 \Rightarrow n = 12$.
Q12 — Summation of Special Series · easy · numerical
The sum $2 + 4 + 6 + \cdots + 2n$ for $n = 25$ is:
A. $650$ ✓ Correct
B. $625$
C. $675$
D. $600$
Solution: $= n(n+1) = 25 \cdot 26 = 650$.
Q13 — Summation of Special Series · hard · numerical
The value of $\sum_{r=11}^{15} r^2$ (i.e. $\sum_{1}^{15} - \sum_{1}^{10}$) is:
A. $620$
B. $865$
C. $385$
D. $855$ ✓ Correct
Solution: $\sum_{1}^{15} r^2 = 1240$ and $\sum_{1}^{10} r^2 = 385$; difference $= 855$.
Q14 — Summation of Special Series · medium · numerical
The sum $1 \cdot 2 + 2 \cdot 3 + \cdots + 10 \cdot 11$ is:
A. $495$
B. $440$ ✓ Correct
C. $400$
D. $385$
Solution: $\sum r(r+1) = \dfrac{10 \cdot 11 \cdot 12}{3} = 440$.
Q15 — Summation of Special Series · hard · numerical
The value of $\sum_{r=1}^{n} (3r^2 + 2r)$ for $n = 5$ is:
A. $165$
B. $195$ ✓ Correct
C. $180$
D. $210$
Solution: $3 \sum r^2 + 2 \sum r = 3(55) + 2(15) = 165 + 30 = 195$.
Q16 — Summation of Special Series · medium · numerical
The value of $\sum_{r=1}^{4} (r+1)^2$ is:
A. $60$
B. $45$
C. $50$
D. $54$ ✓ Correct
Solution: $2^2 + 3^2 + 4^2 + 5^2 = 4 + 9 + 16 + 25 = 54$.
Q17 — Summation of Special Series · hard · numerical
The sum $1 + 3 + 6 + 10 + \cdots$ (first $5$ triangular numbers) is:
A. $30$
B. $35$ ✓ Correct
C. $40$
D. $45$
Solution: $1 + 3 + 6 + 10 + 15 = 35$, matching $\dfrac{n(n+1)(n+2)}{6} = \dfrac{5 \cdot 6 \cdot 7}{6}$.
Q18 — Summation of Special Series · medium · numerical
The value of $\sum_{r=1}^{4} (r^3 - r)$ is:
A. $100$
B. $90$ ✓ Correct
C. $96$
D. $80$
Solution: $\sum r^3 - \sum r = 100 - 10 = 90$.
Q19 — Summation of Special Series · hard · numerical
The sum of the cubes $1^3 + 2^3 + \cdots + 10^3$ equals the square of:
A. $100$
B. $55$ ✓ Correct
C. $45$
D. $50$
Solution: $\left[\dfrac{10 \cdot 11}{2}\right]^2 = 55^2 = 3025$, so it is the square of $55$.
Q20 — Summation of Special Series · medium · theory
The identity $\sum_{r=1}^{n} r^3 = \left(\sum_{r=1}^{n} r\right)^2$ shows that the sum of the first $n$ cubes equals:
A. the sum of the first $n$ squares
B. twice the sum of the first $n$ numbers
C. the square of the sum of the first $n$ natural numbers ✓ Correct
D. the cube of $n$
Solution: $\sum r^3 = \left[\tfrac{n(n+1)}{2}\right]^2 = \left(\sum r\right)^2$.