Question 60460
first series is 1,2,3,4,5,6,7,..........
this is arithmatic series
here the first term is 1 and the common difference is i

now   N th term of an arithmatic progression is given as  a+(N-1)d,  where a is the first term,N is the  Nth term and d is the common difference.


so here the Nth term is   1 + ( N - 1 ) * 1 = 1 + N - 1 = 
understood?
good
now next problem

The series is 1,2,4,8,6,........

this is also an arithmaticseries

here the first term is 1 and the common difference is 2

so the Nth term is given as,

  1 + ( N - 1 ) * 2 = 1 + 2N -2
           = 2N - 1
so for this series Nth term is 2N-1.
understood?


Next series is  1,-1,1,-1,1,-1,........

 this series is a geometric series and here the first term is 1 and common ratio is -1

here the N th term is given by the formula, 

An = a * r ^(n-1), where 'a' is the  first term and 'r ' is the common ratio and ' n' represents  nth  term

so here the nth term is = 1 * ( -1 ) ^ (n-1)
                       
here two possibilities are there,
one is ,  if  n  is an even number then ( n -1) is an odd number.
which means that (-1) ^( an odd number ) is always -1

if n is an odd number then (n-1) is an even number,  which means (-1)^(an even number) is always an even number.

so we can conclude this answer like this,
 if n is an even number, for the given series the nth term is -1
and if n is an odd number the nth term of the series is +1.

understood?


-1,4,-9,16,-25,...
here terms come like this 
first term is -1,  we can put it as -1*(1)^2

second term is 4,  ==> - 1*(2)^2

third term is -9, that is,  -1*(3)^2 

and so on......
so we can write the nth term as,    -1(n)^2
clear?
o.k.


next one is g
 
 1+3+9+27+81+...+.........

first we have to identify the series.

here  the terms are like this

first term is  1

second term is 3, which can be written as 1*3

third term is 9, ===>  1*3^2

fourth term is 27=1*3^3

and so on

so this is a geometric series.

the sum of nth term of a series given by the formula,
(Otherwise we can make out this from the above formation)

Sn = a * r ^ (n-1)
   = 1 * 3 ^(n-1)
   = 1 * 3 ^ (n-1).
clear?
o.k.?
now your doubt is,

1+3+9+27+81+...+1*3^101

 am i correct in assuming that 1*3 is the general term of the sequence and that the last term is the 101st 


here  1 * 3 ^101 MEANS this is the 100 th term of the series.

that means 1*3^(n-1) is the general term of the series.

o.k.?

next one is,    
2+7+12+17+...+97


here the first term is 2

second term is 7= 2+5

third term is  12= 7+5

fourth term is 17=12+5
and so on......

thus the series forms by adding  5 to the previous term.

that is this one is an arithmatic series

here sum of  first n terms of a series is given by the formulais,
n/2(first term +last term)

hwere the first term is  2 and the laqst term is 97so the sum of  the seris is 
n/2(2+97)

for calculating this sum, we need number of terms. right?

for that first of find out the number of terms in the series

we have the formula, a+(n-1)d for the nth term, here take 97 as the nth term

which comes like this,
a + ( n - 1 ) d = 97
2 + ( n -1) * 5 = 97

now subtract 2 from bopth sides of the equation
==> (n-1) * 5 = 95

divide by 5 on both sides

==> (n-1) = 95 / 5
==> n -1 = 19

add 1 on both sides of the equation
==> n =20

so the number of terms in the given series  is 20.  right?

so sum of the 20 terms of the given series is,

20 /2 (2 + 97 )

10 * 99 = 990

so the answer is 990.