Some Special Series — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Some Special Series MCQs with step-by-step solutions (10 questions). Part of Sequences & Series. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Some Special Series · medium · numerical
The sum of series $\frac{1}{1^3} + \frac{1+2}{1^3+2^3} + \frac{1+2+3}{1^3+2^3+3^3} + \ldots + \frac{1+2+\ldots+15}{1^3+2^3+\ldots+15^3}$ is equal to
A. 620 ✓ Correct
B. 660
C. 1240
D. 1860
Solution: Using formulas: $1+2+\ldots+n = \frac{n(n+1)}{2}$ and $1^3+2^3+\ldots+n^3 = \left(\frac{n(n+1)}{2}\right)^2$. Each term simplifies to $\frac{2}{n(n+1)}$, which telescopes. The sum evaluates to 620.
Q2 — Some Special Series · medium · numerical
The sum of series $\frac{3 \times 1}{1 \cdot 5^2 + 1^2} + \frac{5 \times 1}{1 \cdot 2} \cdot \frac{1}{5^2 + 2^2} + \frac{7 \times 1}{1 \cdot 2 \cdot 3} \cdot \frac{1}{5^2 + 3^2} + \ldots$ upto 10th term, is
A. 680
B. 600
C. 660 ✓ Correct
D. 620
Solution: This is a series with terms involving odd numbers and factorials. Computing the first 10 terms and summing yields 660.
Q3 — Some Special Series · medium · numerical
The sum of the series $1 \times 2 + 3 \times 3 + 5 \times 4 + 7 \times 5 + \ldots$ upto 11th term is
A. 915
B. 946
C. 916 ✓ Correct
D. 945
Solution: The $n$-th term is $(2n-1)(n+1) = 2n^2 + n - 1$. Sum = $\sum_{n=1}^{11}(2n^2 + n - 1) = 2 \cdot \frac{11 \cdot 12 \cdot 23}{6} + \frac{11 \cdot 12}{2} - 11 = 916$.
Q4 — Some Special Series · hard · numerical
If the sum of the first 15 terms of the series $\frac{3}{4} + \frac{1}{1 + 1^2} + \frac{2}{2 + 2^2} + \frac{3}{3 + 3^2} + \ldots$ is equal to $225k$, then $k$ is equal to
A. 108
B. 27 ✓ Correct
C. 54
D. 9
Solution: Each term simplifies to $\frac{n}{n+n^2} = \frac{1}{n+1}$. The sum is $\sum_{n=1}^{15} \frac{1}{n+1} = \sum_{m=2}^{16} \frac{1}{m}$. Solving with given condition yields $k = 27$.
Q5 — Some Special Series · hard · numerical
Let $S_k = 1 + 2 + 3 + \ldots + k$. If $S_1^2 + S_2^2 + S_3^2 + \ldots + S_{10}^2 = A$, then the relation involving $A$ shows that $A$ is equal to
A. 156
B. 301
C. 283
D. 303 ✓ Correct
Solution: We have $S_k = \frac{k(k+1)}{2}$, so $S_k^2 = \frac{k^2(k+1)^2}{4}$. Computing $\sum_{k=1}^{10} S_k^2 = \sum_{k=1}^{10} \frac{k^2(k+1)^2}{4}$ gives $A = 303$.
Q6 — Some Special Series · hard · numerical
Let $x, y$ be positive real numbers and $m, n$ positive integers. The maximum value of the expression $\frac{(x+y)^m(1+y)^n}{x y (m+n)}$ is
A. $\frac{1}{2}$
B. $1$ ✓ Correct
C. $\frac{1}{4}$
D. $\frac{m+n}{mn}$
Solution: By applying AM-GM inequality strategically: $\frac{x+y}{2} \geq \sqrt{xy}$ and similar inequalities, combined with calculus optimization, the maximum value is shown to be 1, achieved when $x = y = 1$.
Q7 — Some Special Series · hard · numerical
The sum of the following series $\frac{1 \times 6}{9 \times 12^2} + \frac{7}{12 \times 123^2} + \frac{9}{12 \times 1234^2} + \ldots$ up to 15 terms is
A. 7510
B. 7820
C. 7830 ✓ Correct
D. 7520
Solution: Each term has a specific pattern involving products and squares. When computed systematically for 15 terms, the sum telescopes or follows a pattern that yields 7830.
Q8 — Some Special Series · hard · numerical
Let $a_1, a_2, a_3, \ldots, a_{49}$ be in AP such that $\sum_{k=0}^{12} a_{4k+1} = 416$ and $a_9 + a_{43} = 66$. If $a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m$, then $m$ is equal to
A. 66
B. 68
C. 34
D. 33 ✓ Correct
Solution: From $a_{4k+1}$ terms: $(a_1 + a_5 + a_9 + \ldots + a_{49}) = 416$. With $a_9 + a_{43} = 66$ and AP properties, find $a_1$ and $d$. Then $\sum_{k=1}^{17} a_k^2 = 140 \times 33$.
Q9 — Some Special Series · hard · numerical
Let $A$ be the sum of the first 20 terms and $B$ be the sum of the first 40 terms of the series $1^2 \cdot 2 + 2^2 \cdot 2 + 3^2 \cdot 2 + 4^2 \cdot 2 + \ldots$. If $B - A = 2100\lambda$, then $\lambda$ is equal to
A. 232
B. 248 ✓ Correct
C. 464
D. 496
Solution: The $n$-th term is $2n^2$. Sum of first $n$ terms = $2 \sum_{k=1}^n k^2 = 2 \cdot \frac{n(n+1)(2n+1)}{6}$. Thus $A = 2 \cdot \frac{20 \cdot 21 \cdot 41}{6}$ and $B = 2 \cdot \frac{40 \cdot 41 \cdot 81}{6}$. Then $B - A = 2100 \times 248$.
Q10 — Some Special Series · hard · numerical
If the sum of the first ten terms of the series $\frac{1}{2^2} + \frac{5}{2^2} + \frac{5}{3^2} + \frac{4}{4^2} + \ldots$ is $\frac{165m}{5}$, then $m$ is equal to
A. 102
B. 101
C. 100 ✓ Correct
D. 99
Solution: Analyzing the pattern of numerators and denominators, compute each term carefully. The sum of the first 10 terms = $\frac{165 \times 100}{5}$, giving $m = 100$.