JEE Mathematics Principles Of Mathematical Induction MCQs Set 03

Answer: (a) \( \dfrac{(2n)!}{(n!)^2} > \dfrac{4n}{2n+1} \)

Explanation:
1. n = 2: \( \frac{4!}{(2!)^2} = 6 \) and \( \frac{8}{5} = 1.6 \), so the first is larger. n = 3: 20 against 12/7.
2. Induction step: the left side is multiplied by \( \frac{2(2k+1)}{k+1} \), which is at least \( \frac{10}{3} \) for \( k \ge 2 \), while the right side is multiplied by \( \frac{(k+1)(2k+1)}{k(2k+3)} < 2 \). So the inequality persists.

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (d) \( |\sin nx| \le n|\sin x| \)

Explanation:
1. For n = 1 it holds with equality. If \( |\sin kx| \le k|\sin x| \), then \( |\sin(k+1)x| = |\sin kx \cos x + \cos kx \sin x| \le |\sin kx| + |\sin x| \le (k+1)|\sin x| \).
2. The other options fail: for n = 2, x = pi/2, we get \( |\sin nx| = 0 \) while \( n|\sin x| = 2 \), so \( |\sin nx| \ge n|\sin x| \) is false; and for n = 1, \( |\sin nx| < |\sin x| \) is false.

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (a) \( n^2 \)

Explanation:
1. For n = 1 the sum is 1. For n = 2: \( 1 + 2 \cdot \frac32 = 4 \).
2. Both match \( n^2 \), while \( n(n+1) \) gives 2 at n = 1 and \( (n+1)^2 \) gives 4 at n = 1. In general \( \sum k r^{k-1} \) with \( r = 1 + \frac1n \) simplifies to \( n^2 \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (d) \( (n-1)^3 + n^3 \)

Explanation:
1. The n-th bracket has \( 2n-1 \) consecutive numbers from \( (n-1)^2 + 1 \) to \( n^2 \).
2. Their sum is \( (n^2 - (n-1)^2 - 0)\cdot\frac{(n-1)^2 + 1 + n^2}{2} = (2n-1)(n^2 - n + 1) = (n-1)^3 + n^3 \). Check n = 2: 9 = 1 + 8.

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (c) \( \dfrac{n(n+2)}{3} \)

Explanation:
1. The r-th term is \( \frac{r(r+1)(2r+1)/6}{r(r+1)/2} = \frac{2r+1}{3} \).
2. Sum \( = \frac{n(n+1) + n}{3} = \frac{n(n+2)}{3} \). Check n = 1: 1.

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (b) 22.5

Explanation:
1. The numerator is \( \frac{k^2(k+1)^2}{4} \) and the denominator is \( k^2 \), so each term is \( \frac{(k+1)^2}{4} \).
2. Sum \( = \frac{4 + 9 + 16 + 25 + 36}{4} = \frac{90}{4} = 22.5 \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (d) A, D, B, C

Explanation:
1. A = 55, B = 385, C = 3025, D = 2 x 55 = 110.
2. Increasing order: 55 < 110 < 385 < 3025, that is A, D, B, C.

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (c) \( \dfrac{4n}{n+1} \)

Explanation:
1. \( t_r = S_r - S_{r-1} = \frac{r(r+1)(r+2) - (r-1)r(r+1)}{12} = \frac{r(r+1)}{4} \).
2. So \( \frac{1}{t_r} = \frac{4}{r(r+1)} = 4\left(\frac1r - \frac1{r+1}\right) \) and the sum telescopes to \( 4\left(1 - \frac1{n+1}\right) = \frac{4n}{n+1} \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (b) \( \dfrac{n(n+1)(2n+1)}{6} \)

Explanation:
1. The k-th bracket sums to \( k^2 \) (sum of the first k odd numbers).
2. So the total is \( \sum k^2 = \frac{n(n+1)(2n+1)}{6} \). Check n = 2: 1 + 4 = 5 and \( \frac{2 \cdot 3 \cdot 5}{6} = 5 \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

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

Explanation:
1. The k-th term is \( \frac{k(k+1)/2}{k^2(k+1)^2/4} = \frac{2}{k(k+1)} = 2\left(\frac1k - \frac1{k+1}\right) \).
2. The sum telescopes to \( 2\left(1 - \frac1{n+1}\right) = \frac{2n}{n+1} \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (c) 2

Explanation:
1. Modulo 9: \( 8 \equiv -1 \) and \( 62 \equiv -1 \).
2. So \( 8^{2n} - 62^{2n+1} \equiv (-1)^{2n} - (-1)^{2n+1} = 1 + 1 = 2 \pmod 9 \). The remainder is 2.

Teacher's Note:
1. Check n = 1 and n = 2 first; the highest common factor of the first two values is the likely divisor.
2. Prove it by induction: write the (k+1) expression as a multiple of the k expression plus a term known to be divisible.

Answer: (b) \( \dfrac{1}{\sqrt{3n+1}} \)

Explanation:
1. For n = 1 the left side is \( \frac{2}{4} = \frac12 \) and \( \frac{1}{\sqrt{4}} = \frac12 \), so equality holds. The other bounds are smaller (about 0.447, 0.378 and 0.354) and fail at n = 1.
2. For n = 2 the left side is \( \frac{24}{64} = 0.375 \), and \( \frac{1}{\sqrt7} \approx 0.378 \), which is larger, as the inequality needs. Induction shows the bound for all n.

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (a) \( n! < \left(\dfrac{n+1}{2}\right)^n \)

Explanation:
1. n = 2: \( 2! = 2 \) and \( \left(\frac32\right)^2 = 2.25 \), so n! is smaller. n = 3: 6 against 8.
2. By the AM-GM inequality applied to 1, 2, ..., n: \( \sqrt[n]{n!} < \frac{1 + 2 + \dots + n}{n} = \frac{n+1}{2} \) for n > 1.

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (d) \( \dfrac{3n+5}{3n+1} \)

Explanation:
1. Numerator: \( \sum k(k+1)^2 = \frac{n(n+1)(n+2)(3n+5)}{12} \). Denominator: \( \sum k^2(k+1) = \frac{n(n+1)(n+2)(3n+1)}{12} \).
2. The ratio is \( \frac{3n+5}{3n+1} \). Check n = 1: numerator 4, denominator 2, ratio 2 and \( \frac{8}{4} = 2 \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (a) \( \dfrac{7}{81}\left(10^{n+1} - 9n - 10\right) \)

Explanation:
1. n = 1: the sum is 7 and \( \frac{7}{81}(100 - 9 - 10) = 7 \). n = 2: the sum is 84 and \( \frac{7}{81}(1000 - 28) = 84 \).
2. The k-th term is \( \frac79(10^k - 1) \), so the sum is \( \frac79\left[\frac{10(10^n - 1)}{9} - n\right] = \frac{7}{81}(10^{n+1} - 9n - 10) \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (d) \( \dfrac{1 - (n+1)x^n + nx^{n+1}}{(1-x)^2} \)

Explanation:
1. n = 1: the sum is 1 and the formula gives \( \frac{1 - 2x + x^2}{(1-x)^2} = 1 \). n = 2: the sum is 1 + 2x and the formula gives \( \frac{1 - 3x^2 + 2x^3}{(1-x)^2} = 1 + 2x \).
2. Method: differentiate \( 1 + x + \dots + x^n = \frac{1 - x^{n+1}}{1-x} \), or multiply the sum by x and subtract. The other options fail at n = 1.

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (c) \( \dfrac{8n-7}{4} \)

Explanation:
1. \( t_n = S_n - S_{n-1} = \frac{(4n^2 - 3n) - (4(n-1)^2 - 3(n-1))}{4} \).
2. \( = \frac{8n - 4 - 3}{4} = \frac{8n-7}{4} \). Check n = 1: \( S_1 = \frac14 \) and \( t_1 = \frac14 \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (b) \( \dfrac{3^n - 1}{2 \cdot 3^{(n-1)(n+2)/2}} \)

Explanation:
1. Bracket 1 is 1, bracket 2 is \( \frac13 + \frac19 = \frac49 \), bracket 3 is \( \frac1{27} + \frac1{81} + \frac1{243} = \frac{13}{243} \). The n-th bracket is a GP of n terms with first term \( 3^{-n(n-1)/2} \) and ratio \( \frac13 \).
2. Sum \( = 3^{-n(n-1)/2}\cdot\frac{1 - 3^{-n}}{1 - \frac13} = \frac{3^n - 1}{2 \cdot 3^{n-1+n(n-1)/2}} = \frac{3^n - 1}{2 \cdot 3^{(n-1)(n+2)/2}} \). Check n = 2: \( \frac{8}{2 \cdot 3^2} = \frac49 \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (d) \( a(100) < 100 \)

Explanation:
1. By induction, \( \frac n2 < a(n) < n \) for every \( n > 1 \): n = 2 gives \( a(2) = 1 + \frac12 + \frac13 = \frac{11}{6} \), between 1 and 2.
2. Hence a(100) < 100 is true. The others fail: a(100) > 100 and a(20) = 100 contradict \( a(n) < n \), and a(200) > 100 (because \( a(n) > \frac n2 \)).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

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

Explanation:
1. Check n = 1: \( H_1 = 1 \) and \( 2H_1 - 1 = 1 \). Check n = 2: \( 1 + \frac32 = \frac52 \) and \( 3 \cdot \frac32 - 2 = \frac52 \).
2. Induction step: add \( H_{k+1} \) to \( (k+1)H_k - k \): \( (k+1)H_k - k + H_k + \frac{1}{k+1} = (k+2)H_{k+1} - (k+1) \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (b) \( \dfrac{n}{24}\left(2n^2 + 9n + 13\right) \)

Explanation:
1. \( S_k = \frac{k+1}{2} \), so \( S_k^2 = \frac{(k+1)^2}{4} \). For n = 1 the sum is 1 and the formula gives \( \frac{24}{24} = 1 \).
2. Sum \( = \frac14\left[\frac{(n+1)(n+2)(2n+3)}{6} - 1\right] = \frac{n(2n^2 + 9n + 13)}{24} \). Check n = 2: the sum is \( 1 + \frac94 = \frac{13}{4} \) and the formula gives \( \frac{13}{4} \), while the other options give \( \frac14 \), \( \frac{13}{2} \) and \( \frac{13}{2} \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.

Answer: (a) \( \dfrac{4006}{3009} \)

Explanation:
1. \( \frac1{t_n} = \frac{4}{(n+2)(n+3)} = 4\left(\frac1{n+2} - \frac1{n+3}\right) \), so the sum telescopes.
2. Sum \( = 4\left(\frac13 - \frac1{2006}\right) = 4 \cdot \frac{2003}{6018} = \frac{4006}{3009} \).

Teacher's Note:
1. Test n = 1 and n = 2 in the formula before choosing, then confirm with the induction step (assume true for n = k, prove for n = k + 1).
2. Telescoping and standard sums such as the sum of r, r squared and r cubed solve most series questions quickly.