Recall from my proposition
Sn = n/(n − 1) ∫ Tm dm,
1 ≤ m ≤ m
Where Tm = mth term of any linear progression
Sm = sum of the mth term of the progression
Worked Example
Evaluate
1 + 5 + 9 + ... + (4m − 3)
Here Tm = 4m − 3
Sm = m/(m − 1) ∫ (4m − 3) dm,
1 ≤ m ≤ m
Sm = m/(m − 1) x [(2m² − 3m) − (2 − 3)]
Sm = m/(m − 1) x [ 2m² − 3m + 1]
Sm = m/(m − 1) x (m − 1)(2m − 1)
Sm = m(2m − 1) ___________ that's the sum
Begin your day with a plan, I started mine with maths.
GM
*By Morawo maxwell*
Share
