Sol. (i) Let S n = 1 + 4 + 10 + 19 + 31 + … + t n

or S n = 1 + 4 + 10 + 19 + 31 + … + t n–1 + t n

on subtracting, we get 0 = 1 + 3 + 6 + 9 + …. + (t n – t n–1 ) – t n

⇒ t n = 1 + (3 + 6 + 9 + …. upto n –1 terms) = 1 + [6 + (n – 2) 3] =

Therefore, S n = – = – + n

= – n (n + 1) + n

= [(n +1) (2n) + 1 – 3 (n+1) + 4] =

(ii) Let us prove a general result in order to evaluate the given sum which is a special case of the general result. We have (if x ≠ 1)

1 + x + x 2 + …. + x n–1 = … (i)

On differentiating (i) we get

1 + 2x + 3x 2 + ….. + (n–1) x n–2 = – … (ii)

Now,

1 + 3x + 5x 2 +…. + (2n –1) x n–1

= (1+ x + x 2 + ….. + x n–1 ) + (2x + 4x 2 + 6x 3 + ……+ (2n–1) x n–1 )

= 1 + x + x 2 + ….. + x n–1 + 2x(1 + 2x + 3x 2 +…..+ (n–1) x n–2 )

= + 2x (From (i) and (ii))

Thus 1 + 3x + 5x 2 + ….. + (2n –1) x n–1 = … (iii)

On putting x = ½ in (iii) we get

1 + + + + ….. +

= [3(2 n –1) –2n]

Multiplying both sides by ½ we get

+ + + …….+ = [3(2 n –1) –2n]

which is required sum.