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If S1, S2, S3, … Sm are the sums of n terms of m A.P.,s whose first terms are 1, 2, 3 ... m and whose common differences are 1, 3 ,5, … (2m – 1) respectively, then show that (S1 + S2 + S3 + … + Sm) = (1/2) mn(mn + 1)

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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)

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