Question

The sum of n terms of 1, (1 + 2), (1 + 2 + 3) …….. is (a) (n/3)(n+1)(n–2) (b) (n/3)(n+1)(n+2) (c) n(n+1)(n+2) (d) None​

Answer 1

Answer:

(d) None

Step-by-step explanation:

Solution:

Given series is 1 + (1+2) + (1+2+3)+….

Here nth term = 1+2+3+…+n

Tn = n(n+1)/2

So sum, S = ∑n(n+1)/2

= ½ (∑n2 + ∑n)

= ½ (n(n+1)(2n+1)/6 + n(n+1)/2)

= ½ (n(n+1)/2) [(2n+1)/3 +1)]

= ¼ n(n+1)(2n+4)/3

= n(n+1)(n+2)/6

(d)

Hope it helps you :)

Answer 2

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