Question -
Answer -
Let P (n): a + (a + d) + (a + 2d) + … + (a + (n-1)d) =n/2 [2a + (n-1)d]
Let us check for n = 1,
P (1): a = ½ [2a + (1-1)d]
: a = a
P (n) is true for n = 1.
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): a + (a + d) + (a + 2d) + … + (a + (k-1)d) = k/2 [2a +(k-1)d] … (i)
So,
a + (a + d) + (a + 2d) + … + (a + (k-1)d) + (a + (k)d)
Now, 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
Hence, P (n) is true for all n ∈ N.