JEE Mathematics Sequence and Series MCQs Set 20

Answer: (a) \( \frac{n+1}2 \)

Explanation:
1. The numerator is \( \frac{n(n+1)}2 \), so \( T_n=\frac{n+1}2 \).

Teacher's Note:
1. Divide the sum by n.

Answer: (b) \( -1 \)

Explanation:
1. \( a_3=1,\ a_4=0,\ a_5=-1 \).

Teacher's Note:
1. Apply the recurrence step by step.

Answer: (b) \( 6 \)

Explanation:
1. \( n^3-n=(n-1)n(n+1) \), a product of three consecutive integers.
2. It is always divisible by 6 (but not always by 4 or 9).

Teacher's Note:
1. Factorise n cubed minus n.

Answer: (b) \( 2^{10} \)

Explanation:
1. The values \( 2^0,2^1,\dots,2^9 \) occupy \( 1+2+\dots+512=1023 \) places.
2. So the 1024th and 1025th terms are \( 2^{10} \).

Teacher's Note:
1. Count the cumulative number of terms.

Answer: (c) \( (n^2-n+1)+(n^2-n+3)+(n^2-n+5)+\dots+(n^2+n-1) \)

Explanation:
1. The first number in the group for \( n^3 \) comes after \( 1+2+\dots+(n-1)=\frac{n(n-1)}2 \) odd numbers, so it is \( 2\cdot\frac{n(n-1)}2+1=n^2-n+1 \).
2. The group has n odd numbers, ending at \( n^2-n+1+2(n-1)=n^2+n-1 \). For example n = 3: 7 + 9 + 11.
3. Option (a) gives 3 for n = 2 but then 9 as the second term, so it fails.

Teacher's Note:
1. Test the option with n = 2 and n = 3.

Answer: (d) \( \frac{n(n+1)(n+2)}6 \)

Explanation:
1. The \( k^{th} \) term is \( \frac{k(k+1)}2 \).
2. The sum is \( \frac{n(n+1)(n+2)}6 \).

Teacher's Note:
1. Sum of triangular numbers.

Answer: (b) \( \frac{2n(n+1)(2n+1)}3 \)

Explanation:
1. \( 4\sum k^2=\frac{4n(n+1)(2n+1)}6=\frac{2n(n+1)(2n+1)}3 \).

Teacher's Note:
1. Factor 4 out of the squares.

Answer: (c) \( 2485 \)

Explanation:
1. \( \sum_1^{20}k^2=2870 \) and \( \sum_1^{10}k^2=385 \).
2. The difference is 2485.

Teacher's Note:
1. Subtract the first ten squares.

Answer: (a) \( \frac{234}{25} \)

Explanation:
1. The numerator is \( 78^2=6084 \) and the denominator is \( \frac{12\cdot13\cdot25}6=650 \).
2. The ratio is \( \frac{6084}{650}=\frac{234}{25} \).

Teacher's Note:
1. Use the standard formulas.

Answer: (b) \( 10 \)

Explanation:
1. \( \frac{n(n+1)(2n+1)}6-\frac{n(n+1)}2=\frac{n(n+1)(n-1)}3=330 \).
2. \( n^3-n=990 \), so \( n=10 \).

Teacher's Note:
1. Subtract and simplify.

Answer: (b) \( \frac n{n+1} \)

Explanation:
1. Each term is \( \frac1k-\frac1{k+1} \), so the sum telescopes to \( 1-\frac1{n+1}=\frac n{n+1} \).

Teacher's Note:
1. Use partial fractions.

Answer: (a) \( \frac1{24}n(n-1)(n+1)(3n+2) \)

Explanation:
1. The sum is \( \frac12\left[\left(\sum k\right)^2-\sum k^2\right] \).
2. \( =\frac12\left[\frac{n^2(n+1)^2}4-\frac{n(n+1)(2n+1)}6\right]=\frac{n(n+1)(n-1)(3n+2)}{24} \).

Teacher's Note:
1. Use (sum) squared minus the sum of squares, over 2.

Answer: (d) \( \frac{n(n+1)(9n^2+23n+13)}6 \)

Explanation:
1. The \( k^{th} \) term is \( k(2k+1)(3k+2)=6k^3+7k^2+2k \).
2. \( S=\frac{3n^2(n+1)^2}2+\frac{7n(n+1)(2n+1)}6+n(n+1)=\frac{n(n+1)(9n^2+23n+13)}6 \).
3. Check n = 1: 15.

Teacher's Note:
1. Check with n = 1.

Answer: (b) \( \frac12n^2(n+1) \)

Explanation:
1. For odd n, \( S_n=S_{n+1}-2(n+1)^2 \) where \( n+1 \) is even.
2. \( =\frac{(n+1)(n+2)^2}2-2(n+1)^2=\frac{n+1}2\cdot n^2=\frac12n^2(n+1) \).

Teacher's Note:
1. Add one more term to reach an even count.

Answer: (c) \( \frac n2\log\left(\frac{a^{n+1}}{b^{n-1}}\right) \)

Explanation:
1. The sum is \( \log a\sum r-\log b\sum(r-1)=\frac{n(n+1)}2\log a-\frac{n(n-1)}2\log b \).
2. \( =\frac n2\log\frac{a^{n+1}}{b^{n-1}} \).

Teacher's Note:
1. Split the logarithm.

Answer: (d) \( \frac14(n+1)^2(2n-1) \)

Explanation:
1. For \( n=1 \) the value is 1 and for \( n=3 \) it is \( 27-8+1=20 \).
2. Only \( \frac14(n+1)^2(2n-1) \) gives 1 and 20.

Teacher's Note:
1. Test small odd n.

Answer: (d) \( \frac1{72} \)

Explanation:
1. The \( k^{th} \) term is \( \frac{4k+1}{(4k-1)^2(4k+3)^2}=\frac18\left[\frac1{(4k-1)^2}-\frac1{(4k+3)^2}\right] \).
2. The series telescopes to \( \frac18\cdot\frac1{3^2}=\frac1{72} \).

Teacher's Note:
1. Use (4k + 3) squared minus (4k - 1) squared = 8(4k + 1).

Answer: (a) \( (1+x)/(1-x)^3 \)

Explanation:
1. The series is \( \sum k^2x^{k-1}=\frac{1+x}{(1-x)^3} \) for \( |x|\lt 1 \).

Teacher's Note:
1. It is derived by differentiating x/(1 - x)^2.

Answer: (c) \( \frac12\left(\sum_{r=1}^nr^2-\sum_{r=1}^nr\right) \)

Explanation:
1. \( \sum_{m=1}^n\frac{m(m+1)}2=\frac12\left(\sum m^2+\sum m\right) \).
2. The difference is \( \sum r^2-\frac12\sum r^2-\frac12\sum r=\frac12\left(\sum r^2-\sum r\right) \).

Teacher's Note:
1. The double sum is a sum of triangular numbers.

Answer: (a) \( n(n+1)(n+2) \)

Explanation:
1. \( 3\sum k(k+1)=3\cdot\frac{n(n+1)(n+2)}3=n(n+1)(n+2) \).

Teacher's Note:
1. Use sum of k(k + 1).

Answer: (d) \( \frac{2(n+1)}{n+2} \)

Explanation:
1. The \( k^{th} \) term is \( \frac2{k(k+1)}=2\left(\frac1k-\frac1{k+1}\right) \).
2. For \( n+1 \) terms: \( 2\left(1-\frac1{n+2}\right)=\frac{2(n+1)}{n+2} \).

Teacher's Note:
1. Telescope using partial fractions.

Answer: (c) \( (n+1)!-1 \)

Explanation:
1. \( k\cdot k!=(k+1)!-k! \), so the sum telescopes to \( (n+1)!-1 \).

Teacher's Note:
1. Write k as (k + 1) - 1.

Answer: (c) \( \frac{6n}{n+1} \)

Explanation:
1. The \( k^{th} \) term is \( \frac{2k+1}{k(k+1)(2k+1)/6}=\frac6{k(k+1)} \).
2. The sum is \( 6\left(1-\frac1{n+1}\right)=\frac{6n}{n+1} \).

Teacher's Note:
1. Cancel the factor (2k + 1).

Answer: (b) an odd integer divisible by 5

Explanation:
1. The sum is \( 210^2-55^2=44100-3025=41075 \).
2. It ends in 5, so it is odd and divisible by 5.

Teacher's Note:
1. Use the square of the triangular number.

Answer: (b) \( 53261 \)

Explanation:
1. The cubes are \( 5^3=125 \) up to \( 21^3=9261 \) (since \( 22^3=10648>10000 \)).
2. \( \sum_5^{21}k^3=231^2-10^2=53361-100=53261 \).

Teacher's Note:
1. Find the first and last cube in range.