Question

Hey!! how y'all doing? I need the answer of this question. (21 pts each)

Answer 1

Heya!!!

Kindly refer to the attachment for solution ^_^

Hope it helps ^_^

Answer 2

Answer:

S₁ + S₃ = 2S₂

Step-by-step explanation:

Sum of n terms of an AP sn = (n/2)[2a + (n - 1) * d]

(i) S₁

First term(a) = 1

Common difference(d) = 1

Sum of n terms = (n/2)[2 + (n - 1) * 1]

                         = (n/2)[2 + n - 1]

                         = (n/2)[n + 1]

                         = n(n+1)/2

(ii) S₂:

First term(a) = 1

Common difference(d) = 2

Sum of n terms = (n/2)[2a + (n - 1) * d]

                          = (n/2)[2 + (n - 1) * 2]

                          = (n/2)[2 + 2n - 2]

                          = (n/2)[2n]

                          = n²

(iii) S₃:

First term(a) = 1

Common difference (d) = 3

Sum of n terms = (n/2)[2a + (n - 1) * d]

                          = (n/2)[2 + (n - 1) * 3]

                          = (n/2)[2 + 3n - 3]

                          = (n/2)[3n - 1]

                          = n(3n - 1)/2

LHS:

S₁ + S₃ = {n(n+1)/2} + {n(3n - 1)/2}

           = n[n + 1 + 3n - 1]/2

           = n[n + 3n]/2

           = n[4n]/2

           = 2n²

           = 2S₂ {From (ii)}

           = RHS

Therefore, S₁ + S₃ = 2S₂

Hope it helps!