Mathematical InductionAnswer all questions and show all workings.Question 1 Prove the following...

Question

Mathematical InductionAnswer all questions and show all workings.Question 1 Prove the following preposition for all positive integers n.Question 2

Nitika · Accepted Answer

Solutions
Solution 1 (i). Let P(n): 101+102+103+…+10n = 
Base case		Put n=1 in P(n)
L.H.S: 101=10
R.H.S: 
Hence P(n) is true for n=1
Inductive hypothesis	Let P(n) is true for P(k), k is positive integer
101+102+103+…+10k = 				equation (1)
Inductive step	If P(n) is true for n=k then P(n) must also be true for n=k+1
			Put n=k+1 in P(n)
			L.H.S=101+102+103+…+10k+1
=101+102+103+…+10k+1
=101+102+103+…+10k+10k+1
Using equation 1
=	+10k+1
=
Taking  common, we get
=
=
=	
=R.H.S
Since the statement is true for n=1 and n=k+1, by the principal of mathematical induction p(n) is true for all positive integers n.
Solution 1(ii)			Let P(n): 
Base case			Put n=1 in P(n)	Add the values by varying r from 1 to n
L.H.S= 1(1+1)=2	For n=1: r can take only one value i.e.

Mathematical Induction Answer all questions and show all workings. Question 1 Prove the following preposition for all positive integers n. Question 2

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