Easy one I guess, but i want to see the method. The odd numbers are arranged as shown below:
(1)
(3,5)
(7,9,11)
(13,15,17,19)...
Find the sum of terms in the nth bracket as a function of n.
3two Arithmetic progression .
First we have to find the first term in the n th bracket then we need to sum the progression using the formula for sum to n terms.
341i am looking for a convincing proof that the sum of terms in the nth bracket is n3
1SEE IN SHORT I M TELLING U SOLUTION........
OBSERVE FIRST TERM OF EACH BRACKET...............
DID U GET THE PATTERN????
YES I HOPE.........THE FIRST TERM OF nth bracket is( n2-n+1)
and u can see clearly nth bracket has n terms..........
now this is a cakewalk..........
cheers!!!!!!!!
9clue
(n(n+1)/2)2 - (n(n-1)/2)2 = n3
i ll post full solution if this isnt enuf
11the first term of the nth bracket is n2 - n + 1
hence the sum of the n terms of the nth bracket is
= n/2{2(n2 - n + 1) + (n-1)2} (using the form.Sn=n/2{2a+(n-1)d})
=n/2{2n2}
=n3
341What Celestine was hinting at was a more painless solution.
The sum of all numbers from the 1st bracket to the nth bracket is the sum of the 1st n(n+1)/2 odd numbers = Sn = \frac{n^2 (n+1)^2}{4} which is \sum n^3
So if you call the sum of numbers in the nth bracket as Tn, then
\sum T_n = S_n which implies that T_n = n^3 or if u want to be doubly sure , you just say T_n = S_n - S_{n-1} = n^3
