JEE Mathematics Principles Of Mathematical Induction MCQs Set 02

Answer: (a) \( \dfrac{n(4n^2 + 6n - 1)}{3} \)

Explanation:
1. Test n = 1: the sum is 3 and the formula gives \( \frac{1(4 + 6 - 1)}{3} = 3 \). Test n = 2: the sum is 3 + 15 = 18 and the formula gives \( \frac{2(16 + 12 - 1)}{3} = 18 \).
2. The r-th term is \( 4r^2 - 1 \), so the sum is \( 4 \cdot \frac{n(n+1)(2n+1)}{6} - n = \frac{n(4n^2 + 6n - 1)}{3} \).

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) \( \dfrac{n(n+1)(n+5)}{3} \)

Explanation:
1. Test n = 1: the sum is 4 and \( \frac{1 \cdot 2 \cdot 6}{3} = 4 \). Test n = 2: the sum is 4 + 10 = 14 and \( \frac{2 \cdot 3 \cdot 7}{3} = 14 \).
2. Directly, \( \sum k(k+3) = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} = \frac{n(n+1)(n+5)}{3} \). 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) \( 3n^2 + 2n \)

Explanation:
1. Group the terms in pairs: \( (2+3) + (5+6) + (8+9) + \dots \) with pair sums 5, 11, 17, ... in AP with first term 5 and common difference 6.
2. Sum of n pairs \( = \frac{n}{2}[10 + 6(n-1)] = 3n^2 + 2n \). Check: n = 1 gives 5, n = 2 gives 16.

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) \( 3n^2 + n \)

Explanation:
1. The first differences are 10, 16, 22, so the second difference is a constant 6, and the n-th term has the form \( an^2 + bn + c \) with a = 3.
2. Fitting 4, 14, 30 gives \( 3n^2 + n \): n = 1 gives 4, n = 2 gives 14, n = 3 gives 30, n = 4 gives 52.

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) \( n^2 + n + 1 \)

Explanation:
1. The differences 4, 6, 8 increase by 2, so the n-th term is \( n^2 + bn + c \).
2. Using n = 1 and n = 2 gives b = c = 1, so the term is \( n^2 + n + 1 \). Check n = 3: 13, n = 4: 21.

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) \( \dfrac{n(n+1)(n+2)(n+3)}{4} \)

Explanation:
1. Test n = 1: the sum is 6 and \( \frac{1 \cdot 2 \cdot 3 \cdot 4}{4} = 6 \). Test n = 2: the sum is 6 + 24 = 30 and \( \frac{2 \cdot 3 \cdot 4 \cdot 5}{4} = 30 \).
2. Each term can be written as \( \frac{r(r+1)(r+2)(r+3) - (r-1)r(r+1)(r+2)}{4} \), so the sum telescopes to \( \frac{n(n+1)(n+2)(n+3)}{4} \).

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) \( \dfrac{2n^3 + 12n^2 + 10n - 84}{6} \)

Explanation:
1. The r-th term is \( (r+2)(r+5) \) for r = 1 to n - 2. For n = 3 the sum is just 18, and \( \frac{54 + 108 + 30 - 84}{6} = 18 \).
2. For n = 4 the sum is 18 + 28 = 46 and \( \frac{128 + 192 + 40 - 84}{6} = 46 \). The other options do not give 18 at n = 3.

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) \( n \cdot 2^n \)

Explanation:
1. The r-th term is \( (r+1)2^{r-1} \). For n = 1 the sum is 2, and for n = 2 it is 2 + 6 = 8.
2. The formula \( n \cdot 2^n \) gives 2 and 8. The others give 1, 4 and 4 at n = 1, so they are wrong.

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) 9

Explanation:
1. For 1, 2, 3 the sum is 1 + 8 + 27 = 36, and for 2, 3, 4 it is 8 + 27 + 64 = 99. The highest common factor of 36 and 99 is 9.
2. 27, 54 and 99 do not divide 36, so only 9 works. Induction: the sum for k + 1 minus the sum for k equals \( 9(k+1)^2 \), a multiple of 9.

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) 8

Explanation:
1. n = 1: 9 + 7 = 16. n = 2: 81 + 7 = 88. The highest common factor of 16 and 88 is 8.
2. Induction: \( 3^{2(k+1)} + 7 - (3^{2k} + 7) = 8 \cdot 3^{2k} \), a multiple of 8. So the answer is 8.

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: (a) 19

Explanation:
1. n = 1: 175 + 72 = 247 = 13 x 19. n = 2: 4375 + 432 = 4807 = 19 x 253.
2. The highest common factor of 247 and 4807 is 19 (13 does not divide 4807). Induction: with \( f(k) = 7 \cdot 5^{2k} + 12 \cdot 6^k \), \( f(k+1) = 25 f(k) - 13 \cdot 12 \cdot 6^k \), and \( 13 \cdot 12 \cdot 6^k = 156 \cdot 6^k \) is not generally a multiple of 19, so rely on the checks above: 19 divides both values and is the largest common divisor.

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: (d) 16

Explanation:
1. n = 1: 81 - 17 = 64. n = 2: 729 - 25 = 704 = 64 x 11. So the number is a multiple of 64, hence of 16.
2. 704 is not divisible by 128, 256 or 512, so among the given choices the largest divisor is 16.

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: (a) 2

Explanation:
1. For n = 1: \( 8 = K \cdot 1 \cdot 4 \), so K = 2.
2. Check n = 2: \( 8 + 64 = 72 = 2 \cdot 4 \cdot 9 \). In general \( 8 \sum r^3 = 8 \cdot \frac{n^2(n+1)^2}{4} = 2n^2(n+1)^2 \).

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) \( \dfrac{n^2}{(n+1)^2} \)

Explanation:
1. \( a_k = \frac1k - \frac1{k+1} \), so the sum telescopes to \( 1 - \frac1{n+1} = \frac{n}{n+1} \).
2. Squaring gives \( \frac{n^2}{(n+1)^2} \). Check n = 1: \( \frac12 \) squared is \( \frac14 \).

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: (c) \( \dfrac{x}{1+x} \)

Explanation:
1. The series is \( x - x^2 + x^3 - \dots \), a geometric series with first term x and common ratio \( -x \).
2. Its sum to infinity is \( \frac{x}{1 - (-x)} = \frac{x}{1+x} \). Check x = 0.5: 0.5 - 0.25 + 0.125 - ... = 1/3 = 0.5/1.5.

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: (b) \( \dfrac{\cos\left(\frac{n+1}{2}\theta\right)\sin\left(\frac{n\theta}{2}\right)}{\sin\frac{\theta}{2}} \)

Explanation:
1. Multiply and divide by \( 2\sin\frac\theta2 \) and use \( 2\sin\frac\theta2\cos k\theta = \sin\left(k+\frac12\right)\theta - \sin\left(k-\frac12\right)\theta \). The sum telescopes to \( \frac{\sin\left(n+\frac12\right)\theta - \sin\frac\theta2}{2\sin\frac\theta2} \).
2. This simplifies to \( \frac{\cos\frac{(n+1)\theta}{2}\sin\frac{n\theta}{2}}{\sin\frac\theta2} \). For n = 1 it reduces to \( \cos\theta \).

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) \( \dfrac{n(n+1)^2(n+2)}{12} \)

Explanation:
1. For n = 2 the sum is 1 + 5 = 6 and \( \frac{2 \cdot 9 \cdot 4}{12} = 6 \). For n = 1 the sum is 1 and the formula gives \( \frac{1 \cdot 4 \cdot 3}{12} = 1 \).
2. In general \( \sum_{k=1}^{n} \frac{k(k+1)(2k+1)}{6} = \frac{n(n+1)^2(n+2)}{12} \).

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) \( \dfrac{n(n+1)(n+2)}{6} \)

Explanation:
1. \( t_k = \frac{k(k+1)}{2} \), the k-th triangular number.
2. Sum of the first n triangular numbers is \( \frac{n(n+1)(n+2)}{6} \). Check n = 3: 1 + 3 + 6 = 10 and \( \frac{3 \cdot 4 \cdot 5}{6} = 10 \).

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: (b) \( P(k) \) is true

Explanation:
1. If \( P(k) \) holds, then \( k^2 > 100 \), so \( (k+1)^2 = k^2 + 2k + 1 > 100 + 2k + 1 > 100 \), and \( P(k+1) \) is true.
2. \( P(1) \) and \( P(2) \) are false statements (1 and 4 are not greater than 100), so they give no information about \( P(k+1) \).

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: (a) \( \dfrac{n(n+1)(2n+7)}{6} \)

Explanation:
1. \( \sum k(k+2) = \sum k^2 + 2\sum k = \frac{n(n+1)(2n+1)}{6} + n(n+1) \).
2. \( = \frac{n(n+1)(2n+1+6)}{6} = \frac{n(n+1)(2n+7)}{6} \). Check n = 1: 3 and \( \frac{1 \cdot 2 \cdot 9}{6} = 3 \).

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) \( \dfrac14 \)

Explanation:
1. The r-th term is \( \frac{1}{2r(2r+2)} = \frac14\left(\frac1r - \frac1{r+1}\right) \), which telescopes.
2. Sum \( = \frac14\left(1 - \frac1{n+1}\right) = \frac{n}{4(n+1)} \), so \( k = \frac14 \).

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) 5

Explanation:
1. For n = 1: \( 10 + 3 \cdot 64 = 202 \), and \( 202 + x \) must be a multiple of 9. Since \( 202 = 9 \cdot 22 + 4 \), the least positive x is 5.
2. Check n = 2: \( 100 + 768 + 5 = 873 = 9 \cdot 97 \). Check n = 3: \( 1000 + 3072 + 5 = 4077 = 9 \cdot 453 \).

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.