Question

The odd numbers are divided into groups thus (1,3), (5,7,9,11), (13, 15, 17, 19,21,23
Show that the sum of the numbers in the nth group is 4n^3​

Answer 1

Answer:

4n³

Step-by-step explanation:

Group 1  ( 1 , 3)

Group 2 ( 5 , 7 , 9 , 11)

Group 3 ( 13 , 15 , 17 , 19 , 21 , 23)

Group 1 contains 2 odd numbers

then group 2 contains 2 * 2 = 4 odd numbers

Group 3 contains = 2 * 3 = 6 numbers

Group n Contains = 2n  numbers

Number used before n groups are

2 + 4 + 6 +...............................+ 2(n-1)

= ((n-1)/2)(2 + 2(n-1))

= n(n-1)

= n² - n

1st number of nth group is

n² - n + 1  th odd number   & Value = 2(n² - n + 1 ) - 1  = 2n² - 2n + 1

Last Number of group  = n² - n + 2n th  = n² + n th  & Value = 2(n² + n) - 1

= 2n² + 2n - 1

Number of Terms = 2n

Sum = (2n/2)(2n² - 2n + 1 + 2n² + 2n - 1)

= n( 4n²)

= 4n³