nkaur wrote:For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^k+1 (1/2^k). If T is the sum of the first 10 terms in the sequence, then T is
(1) greater than 2
(2) between 1 and 2
(3) between 1/2 and 1
(4) between 1/4 and 1/2
(5) less than 1/4
How do I have to solve this question?
Thank u
a1=(-1)^1+1*(1/2);
a1= 1/2;
a2= (-1)^2+1*(1/2^2);
a2=-1/4;
a3= 1/8;
:
:
:
a10=-1/1024;
i.e. if we observe half the terms are positive and half of the terms are negative;
from here we can solve the question by using two approaches; separate the positive and negative term, and then evaluate their sum by using G.P. or we can combine consecutive terms to form the new series, i'll be using the last mentioned approach;
(1/2-1/4)+(1/8-1/16)+(1/32-1/64)+(1/128-1/256)+(1/512-1/1024);
1/4+1/16+1/64+1/256+1/1024;
now from here; if we see; the term after 1/4 are start diminishing in magnitude as the denominator increases, therefore surely we can conclude that rest of the terms won't contribute much to the answer, we can also verify this using a G.P.
1/4+1/16+1/64+1/256+1/1024; is a g.p. with common ratio of 1/4;
hence its sum is given as 1/4 (1-(1/4)^5)/(1-1/4); which will approximately be 0.33 hence value must lie between 1/4 and 1/2 hence
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