Solved Problems on Principle of Mathematical Induction are shown here to prove Mathematical Induction.
1. Using the principle of mathematical induction, prove that
1² + 2² + 3² + ..... + n² = (1/6){n(n + 1)(2n + 1} for all n ∈ N.
Solution:
Let the given statement be P(n). Then,
P(n): 1² + 2² + 3² + ..... +n² = (1/6){n(n + 1)(2n + 1)}.
Putting n =1 in the given statement, we get
LHS = 1² = 1 and RHS = (1/6) × 1 × 2 × (2 × 1 + 1) = 1.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): 1² + 2² + 3² + ..... + k² = (1/6){k(k + 1)(2k + 1)}.
Now, 1² + 2² + 3² + ......... + k² + (k + 1)²
= (1/6) {k(k + 1)(2k + 1) + (k + 1)²
= (1/6){(k + 1).(k(2k + 1)+6(k + 1))}
= (1/6){(k + 1)(2k² + 7k + 6})
= (1/6){(k + 1)(k + 2)(2k + 3)}
= 1/6{(k + 1)(k + 1 + 1)[2(k + 1) + 1]}
⇒ P(k + 1): 1² + 2² + 3² + ….. + k² + (k+1)²
= (1/6){(k + 1)(k + 1 + 1)[2(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.
2. By using mathematical induction prove that the given equation is true for all positive integers.
1 x 2 + 3 x 4 + 5 x 6 + …. + (2n  1) x 2n = \(\frac{n (n + 1) (4n  1)}{3}\)
Solution:
From the statement formula
When n = 1,
LHS =1 x 2 = 2
RHS = \(\frac{1(1 + 1) (4 x 1  1)}{3}\) = \(\frac{6}{3}\) = 2
Hence it is proved that P (1) is true for the equation.
Now we assume that P (k) is true or 1 x 2 + 3 x 4 + 5 x 6 + …. + (2k  1) x 2k = \(\frac{k(k + 1)(4k  1)}{3}\).
For P(k + 1)
LHS = 1 x 2 + 3 x 4 + 5 x 6 + …. + (2k  1) x 2k + (2(k + 1)  1) x 2(k + 1)
= \(\frac{k(k + 1)(4k  1)}{3}\) + (2(k + 1)  1) x 2(k + 1)
= \(\frac{(k + 1)}{3}\)(4k^{2}  k + 12 k + 6)
= \(\frac{(k + 1) (4k^{2} + 8k + 3k + 6)}{3}\)
= \(\frac{(k + 1)(k + 2)(4k + 3)}{3}\)
= \(\frac{(k + 1)((k + 1) + 1)(4(k + 1)  1)}{3}\) = RHS for P (k+1)
Now it is proved that P (k + 1) is also true for the equation.
So the given statement is true for all positive integers.
Problems on Principle of Mathematical Induction
3. Using the principle of mathematical induction, prove that
1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + n(n + 1) = (1/3){n(n + 1)(n + 2)}.
Solution:
Let the given statement be P(n). Then,
P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + n(n + 1) = (1/3){n(n + 1)(n + 2)}.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + k(k + 1) = (1/3){k(k + 1)(k + 2)}.
Now, 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +...+ k(k + 1) + (k + 1)(k + 2)
= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ....... + k(k + 1)) + (k + 1)(k + 2)
= (1/3) k(k + 1)(k + 2) + (k + 1)(k + 2) [using (i)]
= (1/3) [k(k + 1)(k + 2) + 3(k + 1)(k + 2)
= (1/3){(k + 1)(k + 2)(k + 3)}
⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +......+ (k + 1)(k + 2)
= (1/3){k + 1 )(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
values of ∈ N.
Problems on Principle of Mathematical Induction
4. By using mathematical induction prove that the given equation is true for all positive integers.
2 + 4 + 6 + …. + 2n = n(n+1)
Solution:
From the statement formula
When n = 1 or P (1),
LHS = 2
RHS =1 × 2 = 2
So P (1) is true.
Now we assume that P (k) is true or 2 + 4 + 6 + …. + 2k = k(k + 1).
For P(k + 1),
LHS = 2 + 4 + 6 + …. + 2k + 2(k + 1)
= k(k + 1) + 2(k + 1)
= (k + 1) (k + 2)
= (k + 1) ((k + 1) + 1) = RHS for P(k+1)
Now it is proved that P(k+1) is also true for the equation.
So the given statement is true for all positive integers.
5. Using the principle of mathematical induction, prove that
1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 +.....+ (2n  1)(2n + 1) = (1/3){n(4n² + 6n  1).
Solution:
Let the given statement be P(n). Then,
P(n): 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 +...... + (2n  1)(2n + 1)= (1/3)n(4n² + 6n  1).
When n = 1, LHS = 1 ∙ 3 = 3 and RHS = (1/3) × 1 × (4 × 1² + 6 × 1  1)
= {(1/3) × 1 × 9} = 3.
LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + ….. + (2k  1)(2k + 1) = (1/3){k(4k² + 6k  1) ......(i)
Now,
1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + …….. + (2k  1)(2k + 1) + {2k(k + 1)  1}{2(k + 1) + 1}
= {1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + ………… + (2k  1)(2k + 1)} + (2k + 1)(2k + 3)
= (1/3) k(4k² + 6k  1) + (2k + 1)(2k + 3) [using (i)]
= (1/3) [(4k³ + 6k²  k) + 3(4k² + 8k + 3)]
= (1/3)(4k³ + 18k² + 23k + 9)
= (1/3){(k + 1)(4k² + 14k + 9)}
= (1/3)[k + 1){4k(k + 1) ² + 6(k + 1)  1}]
⇒ P(k + 1): 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + ..... + (2k + 1)(2k + 3)
= (1/3)[(k + 1){4(k + 1)² + 6(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.
More Problems on Principle of Mathematical Induction
6. By using mathematical induction prove that the given equation is true for all positive integers.
2 + 6 + 10 + ….. + (4n  2) = 2n^{2}
Solution:
From the statement formula
When n = 1 or P(1),
LHS = 2
RHS = 2 × 1^{2} = 2
So P(1) is true.
Now we assume that P (k) is true or 2 + 6 + 10 + ….. + (4k  2) = 2k^{2}
For P (k + 1),
LHS = 2 + 6 + 10 + ….. + (4k  2) + (4(k + 1)  2)
= 2k^{2} + (4k + 4  2)
= 2k^{2 }+ 4k + 2
= (k+1)^{2}
= RHS for P(k+1)
Now it is proved that P(k+1) is also true for the equation.
So the given statement is true for all positive integers.
7.
Using the principle of mathematical induction, prove that
1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{n(n + 1)} = n/(n + 1)
Solution:
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.
Problems on Principle of Mathematical Induction
8. Using the principle of mathematical induction, prove that
{1/(3 ∙ 5)} + {1/(5 ∙ 7)} + {1/(7 ∙ 9)} + ….... + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3)}.
Solution:
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.
Problems on Principle of Mathematical Induction
9. By induction prove that 3^{n } 1 is divisible by 2 is true for all positive integers.
Solution:
When n = 1, P(1) = 3^{1}  1 = 2 which is divisible by 2.
So P(1) is true.
Now we assume that P(k) is true or 3^{k}  1 is divisible by 2.
When P(k + 1),
3^{k + 1}  1= 3^{k} x 3  1 = 3^{k} x 3  3 + 2 = 3(3^{k}  1) + 2
As (3^{k}  1) and 2 both are divisible by 2, it is proved that divisible by 2 is true for all positive integers.
10. Using the principle of mathematical induction, prove that
1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} for all n ∈ N.
Solution:
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.
Problems on Principle of Mathematical Induction
11. By induction prove that n^{2 } 3n + 4 is even and it is true for all positive integers.
Solution:
When n = 1, P (1) = 1  3 + 4 = 2 which is an even number.
So P (1) is true.
Now we assume that P (k) is true or k^{2 } 3k + 4 is an even number.
When P (k + 1),
(k + 1)^{2 } 3(k + 1) + 4
= k^{2 }+ 2k + 1  3k + 3 + 4
= k^{2 } 3k + 4 + 2(k + 2)
As k^{2 } 3k + 4 and 2(k + 2) both are even, there sum also will be an even number.
So it is proved that n^{2 } 3n + 4 is even is true for all positive integers.
12. Using the Principle of mathematical induction, prove that
{1  (1/2)}{1  (1/3)}{1  (1/4)} ….... {1  1/(n + 1)} = 1/(n + 1) for all n ∈ N.
Solution:
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.
Problems on Principle of Mathematical Induction
● Mathematical Induction
11 and 12 Grade Math
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