Question

Find the sum of the terms in the nth bracket of {1},{2,3,4},{5,6,7,8,9}...

Answer 1

the Series forms an AP like 1,3,5....
therefore a=1,d=2
n th bracket would contain
=a+(n-1)d
=1+(n-1)2
=1+2n-2
=2n-1
hence nth bracket would contain 2n-1 no. of terms

Answer 2

Answer:

number of terms in nth bracket = 2n - 1

first term in nth bracket =  (n-1)² + 1   = n² -2n + 2

sum of N terms of Arithmetic Series with first term a and common difference d is (notice here that previous n and N are different)

=> (N/2) × (2a + (N- 1)d)

here N = 2n -1

a = n² -2n + 2

d = 1

substituting

((2n-1) /2) × (2(n² -2n + 2)  + (2n -1 -1))

((2n-1) /2) × (2n² - 4n  + 4 +  2n - 2)

((2n-1) /2) × (2n² -2n + 2)

(2n-1) × (n² -n + 1)