(i) 2 n − 2 (2 n + 2 n − 1 − 1) (ii) (n − 1) 3 + n 3

Sol. (i) 1, (2, 3), (4, 5, 6, 7), (8, 9, ......, 15) ....

Number of terms in n th group = 2 n – 1 ; 1 st term in n th group = 2 n – 1

So Sum of terms in n th group = [2.2 n – 1 + (2 n – 1 – 1) 1] = 2 n – 2 [2 n + 2 n – 1 – 1]

(ii) , (2, 3, 4), (5, 6, 7, 8, 9) ; 1 st term in n th group let T n

T n = a + bn + cn 2 ⇒ T 1 = a + b + c = 1 .... (i)

T 2 = a + 2b + 4c = 2 .... (ii)

and T 3 = a + 3b + 9c = 5 .... (iii)

a = 2, b = – 2, c = 1

On solving (i), (ii) and (iii), we get

So Ι st term is T n = (2 – 2n + n 2 ). Number of terms in n th group = (2n – 1)

∴ sum of terms in n th group = [2 (2 – 2n + n 2 ) + 2n – 2]

= (2n – 1) [n 2 – n + 1] = 2n 3 – 2n 2 + 2n – n 2 + n – 1 = n 3 + (n – 1) 3

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