7. Find the sum of each of the geometric series {-2, ½, -1/8, …, 1/37268}

Question

<!--[if !supportLists]-->1.    <!--[endif]-->Find the sum of each of the geometric series   {-2, ½, -1/8, …, 1/37268}

Answer 1

Your question is incorrect, the last term should be -1/32768 and not 1/37268 else it won't be G.P

In this question, a = -2, r = (1/2)/-2 = -1/4, a_n = -1/32768

But we know, a_n = a₁ * r^n-1

So, -1/32768= -2 * (-1/4)^n-1

or 1/65536= (-1/4)^n-1

or (1/4)⁸ = (-1/4)^n-1

or n-1 = 8

or n= 9

Sum of G.P is given by

Sn = a (1 - rⁿ)/(1-r)

Sn = -2[{1 - (-1/4)⁹} / (1 + 1/4)]

Sn = -52429/32768

So the sum is -52429/32768

Answer 2

The common ratio r=-¼ and the first term a=-2. The series in algebraic terms is a, ar, ar², ...

-2, 1/2, -1/8, 1/32, -1/128, 1/512, -1/2048, 1/8192, -1/32768, ... Odd terms are negative, even terms are positive.

So there is no term 1/32768. I have to assume there's an error and T₉ should be -1/32768, not 1/32768.

Also:

1/32768=ar^n-1=(-2)(-¼)^n-1.

-1/65536=(-¼)^n-1; -1/2¹⁶=(-¼)^n-1=(-1/2²)^n-1; when n is odd, (-¼)^n-1=1/2^2n-2; when n is even, (-¼)^n-1=-1/2^2n-2.

(When n=9, T_n=(-2)(-¼)⁸=(-2)/65536=-1/32768; when n=10, T_n=(-2)(-¼)⁹=-(-2)/262144=1/131072.)

1-rⁿ=(1-r)(1+r+r²+r⁴+...+r^n-1), so S_n=a(1+r+r²+r⁴+...+r^n-1)=

a(1-rⁿ)/(1-r); S₉=-2(1-(-¼)⁹)/(1+¼)=-2=-8(1+1/262144)/5

-262145/163840=-52429/32768≅-1.6.