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13 May 2020, 02:23
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
54% (02:40) correct
46%
(02:50)
wrong
based on 167
sessions
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If \(a_1 = 1\) and \(a_{n+1}= 2a_n + 5\), then \(a_{100}\) is equal to ?
A. \(5*2^{99}-6\)
B. \(5*2^{99}+6\)
C. \(6*2^{99}+5\)
D. \(6*2^{99}-5\)
E. \(6*2^{99}-6\)
Are You Up For the Challenge: 700 Level Questions
A. \(5*2^{99}-6\)
B. \(5*2^{99}+6\)
C. \(6*2^{99}+5\)
D. \(6*2^{99}-5\)
E. \(6*2^{99}-6\)
Are You Up For the Challenge: 700 Level Questions
13 May 2020, 02:44
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\(a_{n+1}= 2a_n + 5\)
\(a_{(1+1)}= 2a_1 + 5\)
\(a_{2}= 2*1 + 5\)
\(a_{2}= 7\)
\(a_{(1+1)}= 6*2^{1}-5\).......(1)
\(a_{(2+1)}= 2a_2 + 5\)
\(a_{3}= 2*7 + 5\)
\(a_{3}= 19\)
\(a_{(2+1)}= 6*2^{2}-5\)........(2)
.
.
. and so on
We can generalize \(a_{n+1} = 6*2^n -5 \)
\(a_{100}= 6*2^{99}-5\)
D
\(a_{(1+1)}= 2a_1 + 5\)
\(a_{2}= 2*1 + 5\)
\(a_{2}= 7\)
\(a_{(1+1)}= 6*2^{1}-5\).......(1)
\(a_{(2+1)}= 2a_2 + 5\)
\(a_{3}= 2*7 + 5\)
\(a_{3}= 19\)
\(a_{(2+1)}= 6*2^{2}-5\)........(2)
.
.
. and so on
We can generalize \(a_{n+1} = 6*2^n -5 \)
\(a_{100}= 6*2^{99}-5\)
D
Bunuel
General Discussion
13 May 2020, 22:44
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