oanhnguyen1116 wrote:
Somebody could give me the clear explanation?
I do not understand much about some ideas above.
Thank you.
Hi,
Since there has been some confusion till now on this Q, let me give you a method
one approach that can be useful is--
Quote:
In the sequence a1,a2,a3...an, each term an is defined as an=1/(−n)^n. The sum of the first ten terms of the sequence (a1,a2,a3...a10) must be __________.
A) Less than −1
B) Between −1 and −1/2
C) Between −1/2 and 0
D) Between 0 and 1/2
E) Greater than 1/2
Solution--
\(a_n=\frac{1}{{-n}^n}\)
so
1) \(a_1=\frac{1}{{-1}^1}\)
2) \(a_2=\frac{1}{{-2}^2}\)..
3) \(a_3=\frac{1}{{-3}^3}\)...
4) \(a_4=\frac{1}{{-4}^4}\)..
INFERENCE:-
1)Starting from -1, alternate terms are -ive and positive..
2) Numeric value keeps decreasing as we go up..
so two scenarios--
1) (first term + second term) + (third term+fourth term) +... (ninth+tenth)..
=> \((-1+\frac{1}{4}) + (-\frac{1}{3^3} + \frac{1}{4^4}) ....+ (-\frac{1}{9^9} + \frac{1}{10^10})\)..
\(-\frac{3}{4} + (- ..) + (- ..)=- \frac{3}{4} - .. - ......\)
keeping the inference in mind, each pair will result in a negative qty
so the SUM will be lesser than \(-\frac{3}{4}\)..
2) lets take first term and then pair of 2nd-3rd, 4th-5th.. and so on
first term + (second term+third term)+(fourth term+fifth term)...(eighth+ninth)+tenth..
=>\(-1+ (\frac{1}{4}-\frac{1}{3^3} ) + ( \frac{1}{4^4}-..) .... (\frac{1}{8^8}-\frac{1}{9^9}) + \frac{1}{10^10}\)..
so leaving -1 and \(\frac{1}{10^10}\), all pair will result in a positive value..
so\(-1+\frac{1}{10^10} +\) some positive value + some positive value +...
so SUM has to be >-1..
Now we have two limits for the SUM..
\(SUM<-\frac{3}{4}\)and\(Sum>-1\)..
so it lies between -1 and -3/4..
choice B contains this RANGE..
so ans B. Between −1 and −1/2