Question

The natural numbers are divided into groups
1; 2,3; 4,5,6 ; 7,8,9,10;.....
Find the sum of the numbes in the kth bracket.​

Answer 1

Answer:

sn=n/2(2a+(n-1)d)

1st group=1

2nd group=2

3rd group=3

nth group=n

first terms of the group:-

1st group=1

2nd group=2

3group=4

4 th group=7

nth group=tn

sum of first terms=1+2+4+7+..........+tn-1st eq

s= 1+2+4+7+….....+tn-1+tn-2nd eq

subtract eq 2-eq1

:- 0=1+1+2+3+4+....………(n-1)-tn

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

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

we know that common difference (d)=1

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

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

=n/2(n^2+1)