Prove by the principle of mathematical induction: a + (a + d) + (a +

1.	Prove by the principle of mathematical induction: a + (a + d) + (a + 2d) + … + (a + (n - 1)d) = n/2[2a + (n - 1)d]
Answer» Suppose P (n): a + (a + d) + (a + 2d) + … + (a + (n - 1)d) = n/2[2a + (n - 1)d] Now let us check for n = 1, P (1): a = 1/2[2a + (1 - 1)d] : a = a P (n) is true for n = 1. Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true. P (k): a + (a + d) + (a + 2d) + … + (a + (k - 1)d) = k/2[2a + (k - 1)d] … (i) Therefore, a + (a + d) + (a + 2d) + … + (a + (k - 1)d) + (a + (k)d) Then, substituting the value of P (k) we get, = k/2 [2a + (k - 1)d] + (a + kd) by using equation (i) = [2ka + k(k - 1)d + 2(a + kd)]/2 = [2ka + k²d – kd + 2a + 2kd]/2 = [2ka + 2a + k²d + kd]/2 = [2a(k + 1) + d(k² + k)]/2 = (k + 1)/2 [2a + kd] P (n) is true for n = k + 1 Thus, P (n) is true for all n ∈ N.

Prove by the principle of mathematical induction: a + (a + d) + (a + 2d) + … + (a + (n - 1)d) = n/2[2a + (n - 1)d]

Answer»

Suppose P (n): a + (a + d) + (a + 2d) + … + (a + (n - 1)d) = n/2[2a + (n - 1)d]

Now let us check for n = 1,

P (1): a = 1/2[2a + (1 - 1)d]

: a = a

P (n) is true for n = 1.

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): a + (a + d) + (a + 2d) + … + (a + (k - 1)d) = k/2[2a + (k - 1)d] … (i)

Therefore,

a + (a + d) + (a + 2d) + … + (a + (k - 1)d) + (a + (k)d)

Then, substituting the value of P (k) we get,

= k/2 [2a + (k - 1)d] + (a + kd) by using equation (i)

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

= [2ka + k²d – kd + 2a + 2kd]/2

= [2ka + 2a + k²d + kd]/2

= [2a(k + 1) + d(k² + k)]/2

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

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.