Mathematical Induction Questions :
Here we are going to see some mathematical induction problems with solutions.
Define mathematical induction :
Mathematical Induction is a method or technique of proving mathematical results or theorems
The process of induction involves the following steps.
Question 1 :
Using the Mathematical induction, show that for any natural number n,
1/(2.5) + 1/(5.8) + 1/(8.11) + · · · + 1/(3n − 1)(3n + 2) = n/(6n + 4)
Solution :
Let p(n) = 1/(2.5) + 1/(5.8) + 1/(8.11) + · · · + 1/(3n − 1)(3n + 2) = n/(6n + 4)
Step 1 :
put n = 1
p(1) = p(n) = [1/(3 − 1)(3 + 2) = 1/(6 + 4)
1/2(5) = 1/10
1/10 = 1/10
Hence p(1) is true.
Step 2 :
Let us assume that the statement is true for n = k
1/(2.5) + 1/(5.8) + 1/(8.11) + · · · + 1/(3K − 1)(3K + 2) = K/(6K + 4)
We need to show that P(k + 1) is true. Consider,
Step 3 :
Let us assume that the statement is true for n = k + 1
p(k+1)
1/(2.5) + 1/(5.8) + 1/(8.11) + · · · + 1/(3K + 2)(3K + 5) = (K+1)/(6K+10)
By applying (1) in this step, we get
L.H.S
= (k+1)(3k+2) / 2(3k+2) (3k+5)
= (k+1)/(6k+10) ----->R.H.S
for any natural number n,
1/(2.5) + 1/(5.8) + 1/(8.11) + · · · + 1/(3n − 1)(3n + 2) = n/(6n + 4)
Question 2 :
Prove by Mathematical Induction that
1! + (2 × 2!) + (3 × 3!) + ... + (n × n!) = (n + 1)! − 1.
Solution :
Let p(n) = 1(1!) + 2(2!) + 3(3!) + ... + n(n!) = (n+1)! - 1
Step 1 :
Put n = 1,
p(n) = 1(1!) + 2(2!) + 3(3!) + ... + n(n!)
Then, LHS = 1(1!) = 1 ⋅ 1 = 1
And RHS = (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1
So, both LHS and RHS = 1 and equation is true at n = 1.
Hence p(1) is true.
Step 2 :
put n = k,
The equation is assumed to be true, and is
1(1!) + 2(2!) + 3(3!) + ...... + k(k!) = (k+1)! - 1 ------------ (1)
We need to show that P(k + 1) is true. Consider,
Step 3 :
Let us assume that the statement is true for n = k + 1
put n = k + 1,
1(1!) + 2(2!) + 3(3!) + ...... + k(k!) + (k+1)[(k+1)!] = [(k+1)+1]! - 1
{1(1!) + 2(2!) + 3(3!) + ...... + k(k!)} + (k+1)[(k+1)!] = (k+2)! - 1
{ (k+1)! - 1 } + (k+1)[(k+1)!] = (k+2)! - 1
(k+1)! + (k+1)[(k+1)!] - 1 = (k+2)! - 1
(k+1)! [ 1 + (k+1) ] - 1 = (k+2)! - 1
(k+1)! x (k+2) - 1 = (k+2)! - 1
(k+2)! - 1 = (k+2)! - 1
Hence it is proved.
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