First terms of an A.P are 1, 2, 3,… m
The common difference are 1, 3, 5,… (2m – 1)
By adding (1) (2) (3) we get
S1 + S2 + S3 + … + Sm = (n/2) (n + 1) + (n/2) (3n + 1) + (n/2) (5n + 1) + … + (n/2) [n(2m – 1 + 1)]
= (n/2) [n + 1 + 3n + 1 + 5n + 1 … + n (2m – 1) + m)]
= (n/2) [n + 3n + 5n + … n(2m – 1) + m]
= (n/2) [n (1 + 3 + 5 + … (2m – 1)) + m
= (n/2) [n(m/2) (2m) + m]
= (n/2) [nm2 + m]
S1 + S2 + S3 + … + Sm = (mn/2) [mn + 1]
Hint: 1 + 3 + 5 + ……. + 2m – 1
Sn = (n/2) (a + l)
= (m/2) (1 + 2m -1)
= (m/2) (2m)