Question

Let 'm' be a positive integer .The sum of all even numbers between
m²-m+1 and m²+m+1 is
a) m^4 +m
b)m³+m
c)m³+m²
d) n^5 -1
explain

Answer 1

option b
given m is a positive integer
in m place take any number it taken 3
the sum of all even number between m^2-m+1 and m^2+m+1
so between 3^2-3+1 and 3^2+3+1
9-3+1 , 9+3+1
7 , 13
between 7, 13 (8,10,12 are even numbers so add them)
8+10+12=30
now check options
key 1 : m^4+m
3^4+3
81+3
84 so wrong
key 2 : m^3+m
3^3+3
27+3
30
proved so option b

Answer 2

Case: 1) Let m be odd.
Thus, m^2-m+1 and m^2+m+1 both are odd.
Therefore, first even number between them is m^2-m+2 and the last one is m^2+m.
Total count of such numbers = [{(m^2+m)-(m^2-m+2)}÷2]+1 =[m-1]+1 =m.
The terms will be in arithmetic progression.
Therefore, sum =(Count/2){First + Last}
=(m/2){(m^2-m+2)+(m^2+m)}
=(m/2){2m^2+2}
=m{m^2+1}
=m^3+m
Case: 2) Let m be even.
Then, m^2-m+1 and m^2+m+1 both are odd in this case also.
Thus, remaining things boils out to be same as case: 1) above.
Thus, the correct option is b)m^3+m.

Thanks!