JEE Mathematics Principles Of Mathematical Induction MCQs Set 04

Answer: (c) \( \dfrac{(x+a_1)(x+a_2)\dots(x+a_n)}{a_1a_2\dots a_n} \)

Explanation:
1. For n = 1: \( 1 + \frac{x}{a_1} = \frac{x + a_1}{a_1} \). Assume the result for n = k.
2. Adding the next term \( \frac{x(x+a_1)\dots(x+a_k)}{a_1\dots a_{k+1}} \) to \( \frac{(x+a_1)\dots(x+a_k)}{a_1\dots a_k} \) gives \( \frac{(x+a_1)\dots(x+a_k)(a_{k+1} + x)}{a_1\dots a_{k+1}} \), which is the result for n = 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) \( (x-a)^2 \)

Explanation:
1. Let \( f(x) = x^n - n a^{n-1}x + (n-1)a^n \). Then \( f(a) = a^n - na^n + (n-1)a^n = 0 \) and \( f'(x) = nx^{n-1} - na^{n-1} \), so \( f'(a) = 0 \).
2. A polynomial with a double root at a is divisible by \( (x-a)^2 \). For n = 2: \( x^2 - 2ax + a^2 = (x-a)^2 \).

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) \( \tan^{-1}x - \tan^{-1}\dfrac{x}{n+1} \)

Explanation:
1. \( \frac{x}{k(k+1) + x^2} = \dfrac{\frac xk - \frac x{k+1}}{1 + \frac xk \cdot \frac x{k+1}} \), so each term is \( \tan^{-1}\frac xk - \tan^{-1}\frac x{k+1} \).
2. The sum telescopes to \( \tan^{-1}x - \tan^{-1}\frac{x}{n+1} \). For n = 1 it is the single term \( \tan^{-1}\frac{x}{2 + x^2} \), which matches.

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) \( \tan^{-1}(n+1) - \dfrac\pi4 \)

Explanation:
1. \( \frac{1}{1+k+k^2} = \frac{(k+1) - k}{1 + k(k+1)} \), so each term is \( \tan^{-1}(k+1) - \tan^{-1}k \).
2. The sum telescopes to \( \tan^{-1}(n+1) - \tan^{-1}1 = \tan^{-1}(n+1) - \frac\pi4 \).

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

Explanation:
1. For n = 2: \( 1 + 8 = 9 \) and \( \frac{2 \cdot 9}{2} = 9 \). For n = 4: \( 1 + 8 + 9 + 32 = 50 \) and \( \frac{4 \cdot 25}{2} = 50 \).
2. The other options give 6, 4.5 and 6 at n = 2, so they fail.

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. For n = 1 the sum is 1 and \( \frac{1 \cdot 2}{2} = 1 \). For n = 3: \( 1 + 8 + 9 = 18 \) and \( \frac{9 \cdot 4}{2} = 18 \).
2. At n = 1 the option \( \frac{3n(n+1)}{2} \) gives 3, not 1; at n = 3 the other two options give 12 and 36, not 18. So \( \frac{n^2(n+1)}{2} \) is correct.

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) \( 2n - H_n \)

Explanation:
1. The k-th term is \( \frac{2k-1}{k} = 2 - \frac1k \).
2. Sum \( = 2n - \left(1 + \frac12 + \dots + \frac1n\right) = 2n - H_n \).

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: (b) \( 16\left(2^{15} - 1\right) \)

Explanation:
1. The set \( S_n \) is a GP of n terms with first term n + 1 and ratio 2.
2. Sum \( = (n+1)(2^n - 1) \). For n = 15 this is \( 16(2^{15} - 1) \). Check n = 3: \( 4 \cdot 7 = 28 = 4 + 8 + 16 \).

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: (d) 326

Explanation:
1. \( S_n \) has n terms with numerators in AP: first term n + 1 and common difference n, all over n.
2. Sum \( = (n+1) + \frac{n(n-1)}{2} \). For n = 25: \( 26 + 300 = 326 \). Check n = 4: \( 5 + 6 = 11 \) and \( \frac{5 + 9 + 13 + 17}{4} = 11 \).

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) \( \dfrac{1220}{2^{20}} \)

Explanation:
1. \( S_n \) has n terms: numerators n + 1, n + 2, ..., 2n over \( 2^{n-1} \). For n = 20 the numerators run from 21 to 40.
2. Their sum is \( \frac{20(21 + 40)}{2} = 610 \), so the set sums to \( \frac{610}{2^{19}} = \frac{1220}{2^{20}} \).

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) 2007006

Explanation:
1. Pair the terms: \( (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + \dots + (2001 - 2002)(2001 + 2002) \) gives \( -(1 + 2 + \dots + 2002) \).
2. So \( S = -\frac{2002 \cdot 2003}{2} + 2003^2 = 2003\left(2003 - 1001\right) = 2003 \cdot 1002 = 2007006 \).

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) 513

Explanation:
1. The sum \( \sum_{k=2}^{n} k2^k = (n-1)2^{n+1} \) (check n = 2: 8 = 1 x 8; n = 3: 8 + 24 = 32 = 2 x 16; n = 4: 32 + 64 = 96 = 3 x 32).
2. Setting \( (n-1)2^{n+1} = 2^{n+10} \) gives \( n - 1 = 2^9 = 512 \), so n = 513.

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) 4

Explanation:
1. \( 17 \equiv 2 \pmod 5 \), so \( 17^{30} \equiv 2^{30} \). Since \( 2^4 \equiv 1 \), \( 2^{30} = (2^4)^7 \cdot 2^2 \equiv 4 \).
2. The remainder is 4.

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) 2

Explanation:
1. \( 2^4 = 16 \equiv 1 \pmod 5 \), and \( 301 = 4 \cdot 75 + 1 \).
2. So \( 2^{301} \equiv 2 \pmod 5 \) and 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: (a) 8

Explanation:
1. \( 5^2 = 25 \equiv -1 \pmod{13} \), so \( 5^{99} = 5 \cdot (5^2)^{49} \equiv 5 \cdot (-1)^{49} = -5 \).
2. \( -5 \equiv 13 - 5 = 8 \pmod{13} \). The remainder 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: (d) only (i)

Explanation:
1. (i) is true: \( x + y \) divides \( x^m + y^m \) for odd m, and 2n + 1 is odd.
2. (ii) is false: for n = 1, \( 1 + 2 = 3 \) is not divisible by 6.

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) only (i)

Explanation:
1. (i) is true: \( x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + y^{n-1}) \).
2. (ii) is false: \( x^2 + y^2 \) is not divisible by \( x + y \) (for x = 1, y = 2: 5 is not divisible by 3); the statement is true for odd n.

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) 7

Explanation:
1. 183! contains the factors 2 and 5, so its unit digit is 0.
2. Powers of 3 end in 3, 9, 7, 1 repeatedly, and \( 183 = 4 \cdot 45 + 3 \), so \( 3^{183} \) ends with the same digit as \( 3^3 = 27 \), namely 7. The sum ends in 7.

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) \( A > B > C \)

Explanation:
1. For \( n > 6 \) we have \( \left(\frac n2\right)^n > n! > \left(\frac n3\right)^n \) (proved by induction).
2. With n = 600: \( 300^{600} > 600! > 200^{600} \), so A > 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) \( > \sqrt n \)

Explanation:
1. n = 2: \( 1 + 0.7071 = 1.7071 > \sqrt2 = 1.4142 \). Suppose the sum for k exceeds \( \sqrt k \).
2. Then the sum for k + 1 exceeds \( \sqrt k + \frac1{\sqrt{k+1}} = \frac{\sqrt{k(k+1)} + 1}{\sqrt{k+1}} \), and since \( \sqrt{k(k+1)} > k \) this is greater than \( \frac{k+1}{\sqrt{k+1}} = \sqrt{k+1} \).

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) \( (2r+1)! \)

Explanation:
1. The product equals \( (n-r)(n-r+1)\dots(n-1)\,n\,(n+1)\dots(n+r) \), a product of \( 2r + 1 \) consecutive natural numbers.
2. A product of m consecutive natural numbers is divisible by m!. For n = r + 1 the product is exactly \( (2r+1)! \), so no larger factorial or multiple works.

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) 66

Explanation:
1. The coefficient of \( x^{n-1} \) in \( (x+1)(x+2)\dots(x+n) \) is \( 1 + 2 + \dots + n = \frac{n(n+1)}{2} \) (proved by induction: multiplying by \( x + k + 1 \) adds \( k+1 \) to the coefficient).
2. Here \( x^{10} = x^{11-1} \), so the coefficient is \( \frac{11 \cdot 12}{2} = 66 \).

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) 8000

Explanation:
1. The k-th bracket is \( (k-1)^2 + (k-1)k + k^2 = k^3 - (k-1)^3 \), and the single 1 at the start is the bracket for k = 1.
2. The sum telescopes: \( \sum_{k=1}^{20}\left[k^3 - (k-1)^3\right] = 20^3 = 8000 \).

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) 7

Explanation:
1. Let \( P(k) = k^7 - k \). By the binomial theorem \( (k+1)^7 - (k+1) = (k^7 - k) + 7(k^6 + 3k^5 + \dots + k) \), so if 7 divides \( P(k) \) it divides \( P(k+1) \); and P(1) = 0.
2. Hence \( n^7 - n \) is divisible by 7 for every n, and \( (n+1)^7 - n^7 - 1 = P(n+1) - P(n) \) is divisible by 7. Check: n = 1 gives 126 = 7 x 18 and n = 2 gives 2058 = 7 x 294; 5, 9 and 11 do not divide both values.

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.