Question

Using mathematical induction, prove that a + (a + d) + (a + 2d) + ... upto n terms = $\frac{n}{2} [2a + (n - 1)d]$ for all n ∈ N

Answer 1

Let

S(n): a + (a+d) + (a+2d) + .... + [a+(n-1)d] = n/2[2a + (n-1)d]

CASE 1:

Put n = 1

S(1) :      a = 1/2[2a + (1-1)d]

            a = 1/2[2a + (0)d]

            a = 1/2[2a]

             a = a

The condition I is satisfied

Case 2:

Now suppose that S(n) is true for n=k

S(k): a + (a+d) + (a+2d) + .... + [a+(k-1)d] = k/2[2a + (k-1)d]   ........ (i)

The statement for n = k+1 becomes

S(k+1): a + (a+d) + (a+2d) + .... + [a+((k+1) -1)d] = (k+1)/2[2a + (k+1)-1)d]

          a + (a+d) + (a+2d) + .... + [ a+ kd ] = (k+1)/2[2a + kd]  ........ (ii)

Now considering (i) again, we have

a + (a+d) + (a+2d) + .... + [a+(k-1)d] = k/2[2a + (k-1)d]

Adding   a + kd   on both sides, we have

a + (a+d) + (a+2d) + .... + [a+(k-1)d]  + [a+ kd]= k/2[2a + (k-1)d] + [a+ kd]

⇒  a + (a+d) + (a+2d) + ....  + [a+ kd] = k/2[2a + kd - d] + [a+ kd]

                                                          = 1/2 [ k{2a + kd - d} + 2{a+ kd} ]

                                                           = 1/2 [ 2ak + k^2d - kd + 2a + 2kd ]

                                                           = 1/2 [ 2ak + k^2d + kd + 2a ]

                                                           = 1/2 [ 2ak + 2a + k^2d + kd ]

                                                           = 1/2 [ 2a (k + 1) + kd (k + 1) ]

                                                            = 1/2 [ (k + 1) ( 2a + kd)]

       a + (a+d) + (a+2d) + ....  + [a+ kd]  =  (k + 1)/2 [2a + kd]

Which is same as expression number (ii)

Thus S(k+1) is true if S(k) is true, so Condition II is satisfied.

And

S(n):  a + (a+d) + (a+2d) + .... + [a+(n-1)d] = n/2[2a + (n-1)d]

is true for all n ∈ N.