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

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....