**CBSE Class 11 Maths – MCQ and Online Tests – Unit 4 – Principle of Mathematical Induction**

Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.

These online tests are based on latest CBSE syllabus. While attempting these our students can identify the weak lessons and continuously practice those lessons for attaining high marks. It also helps to revise the NCERT textbooks thoroughly.

**CBSE Class 11 Maths – MCQ and Online Tests – Unit 4 – Principle of Mathematical Induction**

Question 1.

Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing n shots.

(a) n(n+1)(n+2)/3

(b) n(n+1)(n+2)/6

(c) n(n+2)/6

(d) (n+1)(n+2)/6

## Answer

Answer: (b) n(n+1)(n+2)/6

Hint:

Let each side of the base contains n shots,

then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1

= n(n + 1)/2

= (n² + n)/2

Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers

So, Total shots = ?(n² + n)/2

= (1/2) × {?n² + ?n}

= (1/2) × {n(n+1)(2n+1)/6 + n(n+1)/2}

= n(n+1)(n+2)/6

Question 2.

{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =

(a) 1/(n + 1) for all n ? N.

(b) 1/(n + 1) for all n ? R

(c) n/(n + 1) for all n ? N.

(d) n/(n + 1) for all n ? R

## Answer

Answer: (a) 1/(n + 1) for all n ? N.

Hint:

Let the given statement be P(n). Then,

P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).

When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)

Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] · [1 – {1/(k + 2)}]

= [1/(k + 1)] · [{(k + 2 ) – 1}/(k + 2)}]

= [1/(k + 1)] · [(k + 1)/(k + 2)]

= 1/(k + 2)

Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)

? P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ? N.

Question 3.

The nth terms of the series 3 + 7 + 13 + 21 +………. is

(a) 4n – 1

(b) n² + n + 1

(c) none of these

(d) n + 2

## Answer

Answer: (b) n² + n + 1

Hint:

Let S = 3 + 7 + 13 + 21 +……….a_{n-1} + a_{n} …………1

and S = 3 + 7 + 13 + 21 +……….a_{n-1} + a_{n} …………2

Subtract equation 1 and 2, we get

S – S = 3 + (7 + 13 + 21 +……….a_{n-1} + a_{n}) – (3 + 7 + 13 + 21 +……….a_{n-1} + a_{n})

? 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (a_{n} – a_{n-1}) – a_{n}

? 0 = 3 + {4 + 6 + 8 + ……(n-1)terms} – a_{n}

? a_{n} = 3 + {4 + 6 + 8 + ……(n-1)terms}

? a_{n} = 3 + (n – 1)/2 × {2 ×4 + (n – 1 – 1)2}

? a_{n} = 3 + (n – 1)/2 × {8 + (n – 2)2}

? a_{n} = 3 + (n – 1) × {4 + n – 2}

? a_{n} = 3 + (n – 1) × (n + 2)

? a_{n} = 3 + n² + n – 2

? a_{n} = n² + n + 1

So, the nth term is n² + n + 1

Question 4.

n(n + 1)(n + 5) is a multiple of ____ for all n ? N

(a) 2

(b) 3

(c) 5

(d) 7

## Answer

Answer: (b) 3

Hint:

Let P(n) : n(n + 1)(n + 5) is a multiple of 3.

For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.

So, the given statement is true for n = 1, i.e. P(1) is true.

Let P(k) be true. Then,

P(k) : k(k + 1)(k + 5) is a multiple of 3

? K(k + 1)(k + 5) = 3m for some natural number m, … (i)

Now, (k + 1)(k + 2)(k + 6) = (k + 1)(k + 2)k + 6(k + 1)(k + 2)

= k(k + 1)(k + 2) + 6(k + 1)(k + 2)

= k(k + 1)(k + 5 – 3) + 6(k + 1)(k + 2)

= k(k + 1)(k + 5) – 3k(k + 1) + 6(k + 1)(k + 2)

= k(k + 1)(k + 5) + 3(k + 1)(k +4) [on simplification]

= 3m + 3(k + 1 )(k + 4) [using (i)]

= 3[m + (k + 1)(k + 4)], which is a multiple of 3

? P(k + 1) : (k + 1 )(k + 2)(k + 6) is a multiple of 3

? P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ? N.

Question 5.

1/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ….. + 1/{n(n + 1)}

(a) n(n + 1)

(b) n/(n + 1)

(c) 2n/(n + 1)

(d) 3n/(n + 1)

## Answer

Answer: (b) n/(n + 1)

Hint:

Let the given statement be P(n). Then,

P(n): 1/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ….. + 1/{n(n + 1)} = n/(n + 1).

Putting n = 1 in the given statement, we get

LHS = 1/(1 · 2) = and RHS = 1/(1 + 1) = 1/2.

LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): 1/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ….. + 1/{k(k + 1)} = k/(k + 1) ..…(i)

Now 1/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ….. + 1/{k(k + 1)} + 1/{(k + 1)(k + 2)}

[1/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ….. + 1/{k(k + 1)}] + 1/{(k + 1)(k + 2)}

= k/(k + 1)+1/{ (k + 1)(k + 2)}.

{k(k + 2) + 1}/{(k + 1)²/[(k + 1)k + 2)] using …(ii)

= {k(k + 2) + 1}/{(k + 1)(k + 2}

= {(k + 1)² }/{(k + 1)(k + 2)}

= (k + 1)/(k + 2) = (k + 1)/(k + 1 + 1)

? P(k + 1): 1/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ……… + 1/{ k(k + 1)} + 1/{(k + 1)(k + 2)}

= (k + 1)/(k + 1 + 1)

? P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ? N.

Question 6.

The sum of the series 1² + 2² + 3² + ………..n² is

(a) n(n + 1)(2n + 1)

(b) n(n + 1)(2n + 1)/2

(c) n(n + 1)(2n + 1)/3

(d) n(n + 1)(2n + 1)/6

## Answer

Answer: (d) n(n + 1)(2n + 1)/6

Hint:

Given, series is 1² + 2² + 3² + ………..n²

Sum = n(n + 1)(2n + 1)/6

Question 7.

{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =

(a) 1/(n + 1) for all n ? N.

(b) 1/(n + 1) for all n ? R

(c) n/(n + 1) for all n ? N.

(d) n/(n + 1) for all n ? R

## Answer

Answer: (a) 1/(n + 1) for all n ? N.

Hint:

Let the given statement be P(n). Then,

P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).

When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)

Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] · [1 – {1/(k + 2)}]

= [1/(k + 1)] · [{(k + 2 ) – 1}/(k + 2)}]

= [1/(k + 1)] · [(k + 1)/(k + 2)]

= 1/(k + 2)

Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)

? P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ? N.

Question 8.

For any natural number n, 7^{n} – 2^{n} is divisible by

(a) 3

(b) 4

(c) 5

(d) 7

## Answer

Answer: (c) 5

Hint:

Given, 7^{n} – 2^{n}

Let n = 1

7^{n} – 2^{n} = 7^{1} – 2^{1} = 7 – 2 = 5

which is divisible by 5

Let n = 2

7^{n} – 2^{n} = 7^{2} – 2^{2} = 49 – 4 = 45

which is divisible by 5

Let n = 3

7^{n} – 2^{n} = 7^{3} – 2^{3} = 343 – 8 = 335

which is divisible by 5

Hence, for any natural number n, 7^{n} – 2^{n} is divisible by 5

Question 9.

1/(1 · 2 · 3) + 1/(2 · 3 · 4) + …….. + 1/{n(n + 1)(n + 2)} =

(a) {n(n + 3)}/{4(n + 1)(n + 2)}

(b) (n + 3)/{4(n + 1)(n + 2)}

(c) n/{4(n + 1)(n + 2)}

(d) None of these

## Answer

Answer: (a) {n(n + 3)}/{4(n + 1)(n + 2)}

Hint:

Let P (n): 1/(1 · 2 · 3) + 1/(2 · 3 · 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} .

Putting n = 1 in the given statement, we get

LHS = 1/(1 · 2 · 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.

Therefore LHS = RHS.

Thus, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): 1/(1 · 2 · 3) + 1/(2 · 3 · 4) + ……… + 1/{k(k + 1)(k + 2)} = {k(k + 3)}/{4(k + 1)(k + 2)}. ……. (i)

Now, 1/(1 · 2 · 3) + 1/(2 · 3 · 4) + ………….. + 1/{k(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}

= [1/(1 · 2 · 3) + 1/(2 · 3 · 4) + ………..…. + 1/{ k(k + 1)(k + 2}] + 1/{(k + 1)(k + 2)(k + 3)}

= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]

= {k(k + 3)² + 4}/{4(k + 1)(k + 2)(k + 3)}

= (k³ + 6k² + 9k + 4)/{4(k + 1)(k + 2)(k + 3)}

= {(k + 1)(k + 1)(k + 4)}/{4 (k + 1)(k + 2)(k + 3)}

= {(k + 1)(k + 4)}/{4(k + 2)(k + 3)

? P(k + 1): 1/(1 · 2 · 3) + 1/(2 · 3 · 4) + ……….….. + 1/{(k + 1)(k + 2)(k + 3)}

= {(k + 1)(k + 2)}/{4(k + 2)(k + 3)}

? P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ? N.

Question 10.

The sum of the series 1³ + 2³ + 3³ + ………..n³ is

(a) {(n + 1)/2}²

(b) {n/2}²

(c) n(n + 1)/2

(d) {n(n + 1)/2}²

## Answer

Answer: (d) {n(n + 1)/2}²

Hint:

Given, series is 1³ + 2³ + 3³ + ……….. n³

Sum = {n(n + 1)/2}²

Question 11.

If n is an odd positive integer, then an + bn is divisible by :

(a) a² + b²

(b) a + b

(c) a – b

(d) none of

## Answer

Answer: (b) a + b

Hint:

Given number = an + bn

Let n = 1, 3, 5, ……..

an + bn = a + b

an + bn = a³ + b³ = (a + b) × (a² + b² + ab) and so on.

Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..

So, the given number is divisible by (a + b)

Question 12.

Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing n shots.

(a) n(n+1)(n+2)/3

(b) n(n+1)(n+2)/6

(c) n(n+2)/6

(d) (n+1)(n+2)/6

## Answer

Answer: (b) n(n+1)(n+2)/6

Hint:

Let each side of the base contains n shots,

then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1

= n(n + 1)/2

= (n² + n)/2

Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers

So, Total shots = ?(n² + n)/2

= (1/2)×{?n² + ?n}

= (1/2)×{n(n+1)(2n+1)/6 + n(n+1)/2}

= n(n+1)(n+2)/6

Question 13.

For any natural number n, 7^{n} – 2^{n} is divisible by

(a) 3

(b) 4

(c) 5

(d) 7

## Answer

Answer: (c) 5

Hint:

Given, 7^{n} – 2^{n}

Let n = 1

7^{n} – 2^{n} = 7^{1} – 2^{1} = 7 – 2 = 5

which is divisible by 5

Let n = 2

7^{n} – 2^{n} = 7^{2} – 2^{2} = 49 – 4 = 45

which is divisible by 5

Let n = 3

7^{n} – 2^{n} = 7³ – 2³ = 343 – 8 = 335

which is divisible by 5

Hence, for any natural number n, 7^{n} – 2^{n} is divisible by 5

Question 14.

(n² + n) is ____ for all n ? N.

(a) Even

(b) odd

(c) Either even or odd

(d) None of these

## Answer

Answer: (a) Even

Hint:

Let P(n): (n² + n) is even.

For n = 1, the given expression becomes (1² + 1) = 2, which is even.

So, the given statement is true for n = 1, i.e., P(1)is true.

Let P(k) be true. Then,

P(k): (k² + k) is even

? (k² + k) = 2m for some natural number m. ….. (i)

Now, (k + 1)² + (k + 1) = k² + 3k + 2

= (k² + k) + 2(k + 1)

= 2m + 2(k + 1) [using (i)]

= 2[m + (k + 1)], which is clearly even.

Therefore, P(k + 1): (k + 1)² + (k + 1) is even

? P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n)is true for all n ? N.

Question 15.

For all n ? N, 3×5^{2n+1} + 2^{3n+1} is divisible by

(a) 19

(b) 17

(c) 23

(d) 25

## Answer

Answer: (b) 17

Hint:

Given, 3 × 5^{2n+1} + 2^{3n+1}

Let n = 1,

3 × 5^{2×1+1} + 2^{3×1+1} = 3 × 5^{2+1} + 2^{3+1} = 3 × 5³ + 24 = 3 × 125 + 16 = 375 + 16 = 391

Which is divisible by 17

Let n = 2,

3 × 5^{2×2+1} + 2^{3×2+1} = 3 × 5^{4+1} + 2^{6+1} = 3 × 5^{5} + 2^{7} = 3 × 3125 + 128 = 9375 + 128

= 9503

Which is divisible by 17

Hence, For all n ? N, 3 × 5^{2n+1} + 2^{3n+1} is divisible by 17

Question 16.

The sum of the series 1² + 2² + 3² + ………..n² is

(a) n(n + 1)(2n + 1)

(b) n(n + 1)(2n + 1)/2

(c) n(n + 1)(2n + 1)/3

(d) n(n + 1)(2n + 1)/6

## Answer

Answer: (d) n(n + 1)(2n + 1)/6

Hint:

Given, series is 1² + 2² + 3² + ………..n²

Sum = n(n + 1)(2n + 1)/6

Question 17.

{1/(3 · 5)} + {1/(5 · 7)} + {1/(7 · 9)} + ……. + 1/{(2n + 1)(2n + 3)} =

(a) n/(2n + 3)

(b) n/{2(2n + 3)}

(c) n/{3(2n + 3)}

(d) n/{4(2n + 3)}

## Answer

Answer: (c) n/{3(2n + 3)}

Hint:

Let the given statement be P(n). Then,

P(n): {1/(3 · 5) + 1/(5 · 7) + 1/(7 · 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).

Putting n = 1 in the given statement, we get

and LHS = 1/(3 · 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.

LHS = RHS

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): {1/(3 · 5) + 1/(5 · 7) + 1/(7 · 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)

Now, 1/(3 · 5) + 1/(5 · 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3

= {1/(3 · 5) + 1/(5 · 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}

= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]

= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}

= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]

= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}

= (k + 1)/{3(2k + 5)}

= (k + 1)/[3{2(k + 1) + 3}]

= P(k + 1) : 1/(3 · 5) + 1/(5 · 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]

= (k + 1)/{3{2(k + 1) + 3}]

? P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for n ? N.

Question 18.

(1 + x)n = ____ for all n ? N,where x > -1

(a) 1 + nx

(b) 1 – nx

(c) 1 + nx/2

(d) 1 – nx/2

## Answer

Answer: (a) 1 + nx

Hint:

Let P(n): (1 + x) )^{n} = (1 + nx).

For n = 1, we have LHS = (1 + x))^{1} = (1 + x), and

RHS = (1 + 1 · x) = (1 + x).

Therefore LHS = RHS is true.

Thus, P(1) is true.

Let P(k) is true. Then,

P(k): (1 + x)^{1} = (1 + kx). …….. (i)

Now,(1 + x)^{k+1} = (1 + x)^{k} (1 + x)

= (1 + kx)(1 + x) [using (i)]

=1 + (k + 1)x + kx²

= 1 + (k + 1)x + x [Since kx² = 0]

Therefore P(k + 1) : (1 + x)k + 1 = 1 + (k + 1)x

? P(k +1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ? N.

Question 19.

10^{2n-1} + 1 is divisible by ____ for all N ? N

(a) 9

(b) 10

(c) 11

(d) 13

## Answer

Answer: (c) 11

Hint:

Let P (n): (10^{2n-1} + 1) is divisible by 11.

For n=1, the given expression becomes {10^{(2×1-1)} + 1} = 11, which is divisible by 11.

So, the given statement is true for n = 1, i.e., P (1) is true.

Let P(k) be true. Then,

P(k): (10^{2k-1} + 1) is divisible by 11

? (10^{2k-1} + 1) = 11 m for some natural number m.

Now, {10^{2(k-1)-1} – 1 + 1} = (10^{2k+1} + 1) = {10² · 10^{(2k+1)}+ 1}

= 100 × {10^{2k-1} + 1 } – 99

= (100 × 11 m) – 99

= 11 × (100 m – 9), which is divisible by 11

? P (k + 1) : {10^{2(k-1)} – 1 + 1} is divisible by 11

? P (k + 1) is true, whenever P(k) is true.

Thus, P (1) is true and P(k + 1) is true , whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ? N.

Question 20.

For all n?N, 7^{2n} – 48n-1 is divisible by :

(a) 25

(b) 2304

(c) 1234

(d) 26

## Answer

Answer: (b) 2304

Hint:

Given number = 7^{2n} – 48n – 1

Let n = 1, 2 ,3, 4, ……..

7^{2n} – 48n – 1 = 7² – 48 – 1 = 49 – 48 – 1 = 49 – 49 = 0

7^{2n} – 48n – 1 = 7^{4} – 48 × 2 – 1 = 2401 – 96 – 1 = 2401 – 97 = 2304

7^{2n} – 48n – 1 = 7^{6} – 48 × 3 – 1 = 117649 – 144 – 1 = 117649 – 145 = 117504 = 2304 × 51

Since, all these numbers are divisible by 2304 for n = 1, 2, 3,…..

So, the given number is divisible by 2304

## 0 comments:

## Post a Comment