Question

If S₁, S₂ , S₃, …., S_m 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 + ... + S_m = `1/2` mn(mn + 1)

Answer 1

First terms of an A.P. are 1, 2, 3,…. m

The common difference are 1, 3, 5,…. (2m – 1)

S_n = `"n"/2 [2"a" + ("n" - 1)"d"]`

S₁ = `"n"/2 [2 + ("n" - 1)(1)]`

= `"n"/2 [2 + "n" - 1]`

S₁ = `"n"/2 ["n" + 1]`

S₂ = `"n"/2 [2)(2) + ("n" - 1)3]`

= `"n"/2 [4 + 3"n" - 3]`

S₂ = `"n"/2 (3"n" + 1)`   ....(2)

S₃ = `"n"/2 [2(3) + ("n" - 1)5]`

= `"n"/2 [6 + 5"n" - 5]`

= `"n"/2 [5"n" + 1]`  ...(3)

S_m = `"n"/2 [2"m" + ("n" - 1)(2"m" - 1)]`

= `"n"/2 [2"m" + 2"mn" - "n" - 2"m" + 1]`

= `"n"/2 ["n"(2"m" - 1) + 1]`

By adding (1) (2) (3) we get

S₁ + S₂ + S₃ + …… + S_m = `"n"/2 ("n" + 1) + "n"/2 (3"n" + 1)+ "n"/2 (5"n" + 1)  + ….. +  "n"/2 ["n"(2"m"  – 1 + 1)]`

= `"n"/2 ["n" + 1 + 3"n" + 1 + 5"n" + 1  ……. +  "n"(2"m"   – 1) + "m")]`

= `"n"/2 ["n" + 3"n" + 5"n" + …….  "n"(2"m"  – 1) + "m"]`

= `"n"/2 ["n" (1 + 3 + 5 + …… (2"m"  – 1)) + "m"`

= `"n"/2 ["n"("m"/2) (2"m") + "m"]`

= `"n"/2 ["nm"^2 + "m"]`

S₁ + S₂ + S₃ + ……….. + S_m = `"mn"/2 ["mn" + 1]`

Hint:

1 + 3 + 5 + ……. + 2m – 1

S_n = `"n"/2 ("a" + 1)`

= `"m"/2 (1 + 2"m" - 1)`

= `"m"/2 (2"m")`