JEE Mathematics Mathematical Induction MCQs Set 05

Answer: (b) \(\frac{n(n+1)(n+2)(n+3)}{4}\)

Explanation:
1. For \(n=1\) the sum is 6 and \(\frac{1\cdot2\cdot3\cdot4}{4}=6\), so the formula holds.
2. Assume it holds for \(n=k\). Adding the next term \((k+1)(k+2)(k+3)\) gives \((k+1)(k+2)(k+3)\left[\frac k4+1\right]=\frac{(k+1)(k+2)(k+3)(k+4)}{4}\).
3. This is the formula for \(n=k+1\), so it holds for all natural numbers.

Teacher's Note:
1. Check the formula for \( n = 1 \) first; it removes the wrong options at once.
2. In the induction step, factor out the common product.

Answer: (d) \(\left[\frac{n(n+1)}{2}\right]^2\)

Explanation:
1. At \(n=1\) the sum is 1 and \(\left[\frac{1\cdot2}{2}\right]^2=1\).
2. If true for \(k\), then adding \((k+1)^3\) gives \(\frac{(k+1)^2(k^2+4k+4)}{4}=\left[\frac{(k+1)(k+2)}{2}\right]^2\).
3. Hence the formula holds for all \( n \).

Teacher's Note:
1. The sum of the first \( n \) cubes is the square of the sum of the first \( n \) numbers.
2. Test \( n = 2 \): \( 9 = 3^2 \) matches only option \( (d) \).

Answer: (b) \(a-b\)

Explanation:
1. At \(n=1\) the expression is \(a-b\).
2. Assume \(a^k-b^k=(a-b)r\). Then \(a^{k+1}-b^{k+1}=a(a^k-b^k)+b^k(a-b)\).
3. Both terms are divisible by \(a-b\), so the result holds for all \( n \).

Teacher's Note:
1. Write the new expression as a multiple of the old one plus a leftover term.
2. \( a^2 - b^2 \) shows that \( (a+b) \) does not always divide, since \( a^3 - b^3 \) is not a multiple of \( a+b \).

Answer: (a) \(\frac18(2n+1)^2\)

Explanation:
1. At \(n=1\): \(1<\frac98\) is true.
2. If \(1+\dots+k<\frac18(2k+1)^2\), then adding \(k+1\) gives \(<\frac{(2k+1)^2+8(k+1)}{8}=\frac{(2k+3)^2}{8}\) which is \(\frac18[2(k+1)+1]^2\).
3. So the inequality holds for all \( n \).

Teacher's Note:
1. For inequalities, add the next term to both sides and simplify the right side to the target form.
2. Check \( n = 1 \) against option \( (b) \): \( 1 \lt 9/12 \) is false.

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

Explanation:
1. Here \(u_2=3\cdot3-2\cdot2=5=2^2+1\) and \(u_1=2^1+1\).
2. Assume \(u_k=2^k+1\) and \(u_{k-1}=2^{k-1}+1\). Then \(u_{k+1}=3(2^k+1)-2(2^{k-1}+1)=3\cdot2^k-2^k+1=2^{k+1}+1\).
3. So \(u_n=2^n+1\) for all \( n \).

Teacher's Note:
1. Recurrences need two starting values and a strong induction step using two previous terms.
2. Check \( u_0 = 2 = 2^0 + 1 \) as well.

Answer: (b) \(5\)

Explanation:
1. For \(n=4\): \(100>324\) is false.
2. For \(n=5\): \(1000>405\) is true.
3. If it holds for \(k\ge5\), then \(10^{k-1}>810k>81(k+1)\), so it holds for all \(n\ge5\).

Teacher's Note:
1. Find the first \( n \) by direct trial, then prove the step by induction.
2. The induction step uses \( 10k \gt k + 1 \).

Answer: (d) \(2304\)

Explanation:
1. Write \(7^{2n}=49^n=(1+48)^n=1+48n+\binom n2 48^2+\dots\).
2. So \(49^n-48n-1\) is a multiple of \(48^2=2304\).

Teacher's Note:
1. Binomial expansion of \( (1+48)^n \) shows divisibility by 48 squared.
2. At \( n = 1 \) the value is 0, which is consistent.

Answer: (a) \(60\)

Explanation:
1. At \(n=2\) the value is 60, and at \(n=3\) it is 720.
2. Factorise: \(n^2(n^2-1)(n^2+1)\), which is divisible by 3, 4 and 5 for every \( n \).
3. The largest such divisor among the options is 60.

Teacher's Note:
1. Check divisibility by 3, 4 and 5 separately using residues.
2. Since 60 divides 60 and 720, no larger option is possible.

Answer: (b) \(6\)

Explanation:
1. The product of three consecutive integers is divisible by 2 and by 3, hence by 6.
2. At \(n=3\) the value is 6, so nothing larger works.

Teacher's Note:
1. Take the smallest case \( n = 3 \) to cap the divisor.
2. Among three consecutive numbers one is a multiple of 3 and one is even.

Answer: (a) an integer

Explanation:
1. Write it as \(\frac{15n^7+21n^5+70n^3-n}{105}\).
2. At \(n=1\) the numerator is 105, so the value is 1; at \(n=2\) the value is \(\frac{1920+672+560-2}{105}=30\).
3. The numerator is divisible by 3, 5 and 7 for every \( n \) (by Fermat's little theorem), so the value is an integer, but it is not always odd (it is 30 at \( n = 2 \)).

Teacher's Note:
1. Evaluate at \( n = 1 \) and \( n = 2 \) to confirm the answer is an integer and not always odd.
2. Divisibility by the primes 3, 5, 7 is checked using \( n^3 - n, n^5 - n \) and \( n^7 - n \).

Answer: (b) \(\begin{bmatrix}\cos n\theta&\sin n\theta\\-\sin n\theta&\cos n\theta\end{bmatrix}\)

Explanation:
1. Compute \(A^2=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}\).
2. Assume the pattern for \( k \) and multiply by \( A \); the angle-addition formulas give the pattern for \( k+1 \).

Teacher's Note:
1. A rotation matrix raised to \( n \) gives rotation by \( n \) times the angle.
2. Multiplying by \( A \) adds \( \theta \) to the angle.

Answer: (a) \(\begin{bmatrix}1&10x\\0&1\end{bmatrix}\)

Explanation:
1. Compute \(A^2=\begin{bmatrix}1&2x\\0&1\end{bmatrix}\) and \(A^3=\begin{bmatrix}1&3x\\0&1\end{bmatrix}\).
2. By induction \(A^n=\begin{bmatrix}1&nx\\0&1\end{bmatrix}\), so \(A^{10}\) has top right entry \(10x\).

Teacher's Note:
1. Find the pattern in the first few powers and then prove it.
2. Upper triangular matrices with 1 on the diagonal add their off-diagonal entries.

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

Explanation:
1. The nth term is \(S_n-S_{n-1}=\frac{n^2(n+1)-(n-1)^2n}{2}=\frac{n(3n-1)}{2}\).
2. Check: \( n = 1 \) gives \( 1, n = 2 \) gives \( 5, n = 3 \) gives 12.

Teacher's Note:
1. The nth term equals \( S(n) \) minus \( S(n-1) \).
2. Verify with the first three terms.

Answer: (b) \(2\)

Explanation:
1. \( n = 1 \): \(1<1\) is false.
2. \( n = 2 \): \(2<\frac94\) is true.

Teacher's Note:
1. Substitute the smallest values one by one.
2. Equality at \( n = 1 \) is not strict inequality.

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

Explanation:
1. \( n = 2 \): \(2>2\) is false, so \( (b) \) fails.
2. \( n = 3 \): \(6>4\) true. If \(k!>2^{k-1}\) with \(k\ge3\) then \((k+1)!=(k+1)k!>2\cdot2^{k-1}=2^k\).
3. So it holds for all \(n>2\).

Teacher's Note:
1. The base case for an inequality may be larger than 1.
2. Use \( k + 1 \) greater than or equal to 2 in the step.

Answer: (b) \(120\)

Explanation:
1. The product of 5 consecutive integers is divisible by \(5!=120\).
2. At \(n=1\) the product is \(3\cdot4\cdot5\cdot6\cdot7=2520\) and at \(n=6\) it is \(8\cdot9\cdot10\cdot11\cdot12=95040\); the gcd of these is 120, so no larger number works.

Teacher's Note:
1. The product of \( r \) consecutive integers is divisible by \( r \) factorial.
2. Two sample values are enough to cap the divisor.

Answer: (b) \(17\)

Explanation:
1. At \(n=1\): \(3\cdot125+16=391=17\cdot23\).
2. Write \(3\cdot5^{2n+1}=3\cdot5\cdot25^n\) and \(2^{3n+1}=2\cdot8^n\); since \(25\equiv8\pmod{17}\), the sum is \(\equiv17\cdot8^n\equiv0\).
3. Hence it is divisible by 17 for all n (23 divides \( n = 1 \) only).

Teacher's Note:
1. Reduce both powers modulo the divisor to see the cancellation.
2. Check the other options at \( n = 2 \): the value \( 3\cdot 3125+128 = 9503 = 17 \cdot 559 \) is not divisible by 23 or 19.

Answer: (d) \(11\)

Explanation:
1. At \(n=1\): \(2\cdot64+81=209=11\cdot19\).
2. Since \(2\cdot4\cdot16^n+3\cdot27^n\) and \(16\equiv5,\ 27\equiv5\pmod{11}\), it equals \(11\cdot5^n\equiv0\).
3. So it is a multiple of 11 for all \( n \); the value 209 is odd and not divisible by 3 or 9.

Teacher's Note:
1. Test \( n = 1 \) to remove wrong options, then confirm with the congruence.
2. Congruences make the induction step unnecessary.

Answer: (a) \(11\)

Explanation:
1. At \(n=1\) the value is 11.
2. Since \(10\equiv-1\pmod{11}\), \(10^{2n-1}\equiv(-1)^{2n-1}=-1\), so the expression is divisible by 11.

Teacher's Note:
1. Use \( 10 = -1 \) modulo 11 for powers of 10.
2. An odd power of \( -1 \) is \( -1 \).

Answer: (c) \(49\)

Explanation:
1. Write \(8^n=(1+7)^n=1+7n+\binom n2 7^2+\dots\).
2. So \(8^n-7n-1\) is divisible by \(49\).
3. At \(n=2\) the value is 49, which excludes 64, 36 and 25.

Teacher's Note:
1. Expand \( (1+7)^n \) by the binomial theorem.
2. A single value like \( n = 2 \) settles the choice.

Answer: (a) \(8\)

Explanation:
1. At \(n=1\) the value is 8; at \(n=2\) it is 80.
2. Since \(9\equiv1\pmod 8\), \(9^n-1\) is a multiple of 8. 16 fails at \( n = 1 \).

Teacher's Note:
1. Test \( n = 1 \) and \( n = 2 \) to find the common divisor.
2. 9 leaves remainder 1 when divided by 8.

Answer: (b) \(27\)

Explanation:
1. At \(n=1\) the value is 27.
2. Since \(a-b\) divides \(a^n-b^n\), \(41-14=27\) divides it for all \( n \).

Teacher's Note:
1. Use the fact that \( a - b \) divides \( a^n - b^n \).
2. The difference of the bases is the divisor.

Answer: (a) \(1\)

Explanation:
1. By the factor theorem \(x=k\) is a root, so \(k^n-1=0\).
2. For \(k=1\) this holds for every \( n \), and 1 is the least positive integer.

Teacher's Note:
1. Divisibility by \( x - k \) means \( k \) is a root of the polynomial.
2. \( k = 1 \) is the smallest positive integer, and it works.

Answer: (a) \(-1\)

Explanation:
1. Write \(49^n=(1+48)^n=1+48n+64(\dots)\).
2. So \(49^n+16n=1+64n+64(\dots)\equiv1\pmod{64}\).
3. Hence \(k\equiv-1\pmod{64}\), and the least negative integer in the options is \( -1 \).

Teacher's Note:
1. Expand \( 49^n \) as \( (1 + 48)^n \) and keep the first two terms.
2. Terms with \( 48^2 \) are multiples of 64 because \( 48^2 = 2304 = 36 \cdot 64 \).

Answer: (b) \(n\ge3\)

Explanation:
1. The base case is \( n = 3 \) and each case implies the next.
2. So \( P(3), P(4), P(5) \) and so on are all true, that is, for every \(n\ge3\).

Teacher's Note:
1. Induction can start from any base value.
2. The truth begins at the base case and continues upward.

Answer: (d) nothing can be said

Explanation:
1. Since \(n(n+1)\) is always even, \(n^2+n+1\) is always odd, so \( P(n) \) is false for every \( n \).
2. The step \( P(k) \) implies \( P(k+1) \) holds vacuously, but \( P(1) \) is false, so induction proves nothing.

Teacher's Note:
1. Induction needs a true base case.
2. An implication from a false statement is true, but gives no conclusion.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \(\frac{(n+2)!}{(n-1)!}=n(n+1)(n+2)\), a product of three consecutive integers.
2. Such a product contains a multiple of 3 and an even number, so it is divisible by 6. Hence \( A \) is true.
3. \( R \) is the general reason (\( r \) consecutive integers give a multiple of \( r! \)), and it explains \( A \).

Teacher's Note:
1. A product of \( r \) consecutive integers is divisible by \( r \) factorial.
2. Cancel the factorial first to see the consecutive product.