Just use the formula for the sum of n terms of an arithmetic progression. That gives
T1 = 1/2 (1+1) = 1
T2 = 2/2 (2+3) = 5
T3 = 3/2 (4+6) = 15
Now, Tn has n terms, and they start with a=1+n(n-1)/2, ending with a+n-1
So, the rth term in the ith group is i(i-1)/2+r
Tn = n/2 ((1+n(n-1)/2)*2+n-1) = n/2 (n^2+1)
Now you just have to come up with a formula for Sn, the sum of the first n T's. You probably already know the formulas for
∑k and ∑k^3
use them.
Consider the series 1+(2+3)+(4+5+6)+(7+8+9+10)+(11+12+13+14+15)+....
(a) find an expression for the ith term in the rth bracket.
(b) Given that 39×79 is the sum of all the numbers in the first 6 bracket in the series, find the value of K.
1 answer
1 answer