Answer
(a)The nth term is a$_{n}$=a(r)$^{n-1}$=1(1/3)$^{n-1}$
(b)The nth term is a$_{n}$=(-1)$^{n-1}$(1/3)$^{n-1}$
(c)The nth term of the series is $[$(2n-1)/2n$]$ where n starts from 1.
(d)The nth term of the overall series is (n$^{2}$/$\sqrt[n+1]{\pi
}$) where n starts from 1.
Work Step by Step
1(a) The given sequence is 1,1/3,1/9,1/27 and so on.
When we examine this series we find a pattern that the second term=first
term/3.Also the third term =second term/3.The fourth term=third term/3.
The given series is a geometric progression where first term=a=1 and
common ratio=r=1/3
The nth term is a$_{n}$=a(r)$^{n-1}$=1(1/3)$^{n-1}$=(1/3)
$^{n-1}$.
(b)The given sequence is 1,-(1/3),(1/9),-(1/27) and so on.
The given series is an alternating geometric series with the first term=a=1
and common ratio=r=1/3.
The nth term is a$_{n}$=(-1)$^{n-1}$(1/3)$^{n-1}$
(c)The given series is (1/2),(3/4),(5/6),(7/8) and so on.
The numerators of all the terms are in A.P. with a=1 and d=2
The nth term is a+(n-1)d=1+(n-1)2=2n-1
The nth term of the numerators is 2n-1.
The denominators of all the terms are in A.P. with a=2 and d=2
The nth term is a+(n-1)d=2+(n-1)2=2n-2+2=2n.
The nth term of the denominators is 2n.
The nth term of the series is $[$(2n-1)/2n$]$ where n starts from 1.
(d)
The given series is (1/$\sqrt{\pi }$),(4/$\sqrt[3]{\pi }$) and
so on.
The numerators are all perfect squares which follow a pattern n$^{2}$
.
The denominators are in G.P. whose nth term is $\sqrt[n+1]{\pi }$.
So the nth term of the overall series is (n$^{2}$/$\sqrt[n+1]{\pi
}$) where n starts from 1.
