1+ a + a^2 + a^3......?

Answer 1

geometric progression
Sn=1 (a^n)-1)/(a-1)

Answer 2

If a>=1, then clearly it diverges to infinity. If a<=-1, then it just diverges. Otherwise (-1 < a < 1), the sum can be found in this manner:

Call it S. Then aS = a + a^2 + a^3 +..., which is the sum, minus 1. Thus, aS + 1 = S. Solving for S: 1 = S - aS; 1 = (1-a)/S; S = 1-a. That is, as long as |a| < 1.

For more fun, try this method out on the generic a + ax + ax^2 + ax^3 + ... = ?

Answer 3

(1) S = 1 + a + a^2 + a^3 + . . .
(2) aS = a + a^2 + a^3 + . . .

Subtract (which is only valid if abs(a) < 1)

(3) S(1 - a) = 1, therefore S = 1 / (1 - a).

You can see that this is something like correct. If 'a' is really tiny, S is a tiny bit more than 1. For 'a' close to 1, the closer it is, the larger is S.

Answer 4

Let
x = 1+a+a^2+a^3+...
or x = 1+a(1+a^2+a^3+...)
x = 1+ax (since x = 1+a+a^2+a^3+...see above)

Thus
or 1 = x-ax
or 1 = x(1-a)


or x = 1/(1-a)......answer

Answer 5

This is a geometric series with a common ratio of a and a scale factor of 1. The infinite sum is 1 / (1 - a), but only if a has an absolute value less than 1. If a is greater than or equal to 1, the sum is infinite.

Answer 6

thats a geometric progression wil 1 as first term and common ratio as a. it converges only if a lies in [0,1)
Let S=1+a+a^2+a^3...a^n
S*a=a+a^2+a^3+a^4...a^(n+1)
Sa-S=a^(n+1) - 1 rest all terms cancel
so S = [a^(n+1)-1]/(a-1)
when n approaches ∞, the sum converges only if a lies in [0,1)
when a<1 and a≥0, the numerator becomes -1.
so the sum is -1/(a-1) = 1/(1-a)

Answer 7

this has been answered above..... but not sufficiently

let S = 1+a+a^2+..... +a^n .....1
(no need to assume that it is infinte series at this stage)

so aS = a+a^2+a^3+...+a^(n+1)............2

(2)-(1) gives

S(a-1) = a^(n+1)-1 (other terms cancel out)

so S = (a^(n+1)-1) / (a-1) = sum upto a^n

now suppose it is infinite series... then n->infinity

assume a<1; then a^(infinity) = 0; hence

Limit S = (0-1)/(a-1) = 1/(1-a)
n->infinity

Answer 8

Since you have expressed the series as infinite, it represents a power series which will converge to 1/(1-a), when a<1. On other hand, if a>1 then the series diverges to infinity.

Answer 9

it is a g. p. series in which The first term is 1 and the common ratio is is a
by using the sum formula of g.p.
if a>1 then
S =1(a^n - 1) / (a - 1)
if a<1 then
s= 1(1 - a^n) / (1 - a)

Answer 10

Answer: 1/(1-a)
Solution:
x=1+a+a^2+a^3+...
x=1+a(1+a^2+a^3+...)
x=1+ax
x-ax=1
x(1-a)=1
x=1/(1-a)


Hope this solves the problem