What is the nth term of the series 1 + \(\frac{(1+2)}{2}\) + \(\frac{(

1.	What is the nth term of the series 1 + \(\frac{(1+2)}{2}\) + \(\frac{(1+2+3)}{2}\) + .......?(a) \(\frac{(n+1)}{2}\)(b) \(\frac{n(n+1)}{2}\)(c) n2 – (n + 1) (d) \(\frac{(n+1)(2n+3)}{2}\)
Answer» (a) \(\frac{n+1}{2}\) Reqd. nth term = \(\frac{1+2+3+....+n}{n}\) = \(\frac{n(n+1)}{2}\) x \(\frac{1}{n}\) = \(\frac{n+1}{2}\).

What is the nth term of the series 1 + \(\frac{(1+2)}{2}\) + \(\frac{(1+2+3)}{2}\) + .......?(a) \(\frac{(n+1)}{2}\)(b) \(\frac{n(n+1)}{2}\)(c) n2 – (n + 1) (d) \(\frac{(n+1)(2n+3)}{2}\)

Answer»

(a) \(\frac{n+1}{2}\)

Reqd. nth term = \(\frac{1+2+3+....+n}{n}\)

= \(\frac{n(n+1)}{2}\) x \(\frac{1}{n}\) = \(\frac{n+1}{2}\).