Asked by dam
find the nth term of the series 1,3,6,11,31,48.... Hence duduce a formula for calculating the sum of the series
plz help me show working
plz help me show working
Answers
Answered by
Reiny
* = 1st difference , ** = 2nd difference, *** = 3rd difference
# * ** ***
1
3 2
6 3 1
11 5 2 1
31 20 15 13
48 17 -3 -18
You labeled you subject "arithmetic progression", so I expected a linear relationship.
The first, second and third differences are NOT constants, so the relations cannot be linear, quadratic, nor cubic.
I tried to find a pattern in the sums, again, no luck.
sum(1) = 1
sum(2) = 1 + 3 = 4
sum(3) = 4 + 6 = 10
sum(4) = 10 = 11 = 21
sum(5) = 21 + 31 = 52
Even Wolfram could not find a pattern in your terms
http://www.wolframalpha.com/input/?i=pattern+%7B1,+3,+6,+11,+31,+48,+...%7D
I suspect a typo.
# * ** ***
1
3 2
6 3 1
11 5 2 1
31 20 15 13
48 17 -3 -18
You labeled you subject "arithmetic progression", so I expected a linear relationship.
The first, second and third differences are NOT constants, so the relations cannot be linear, quadratic, nor cubic.
I tried to find a pattern in the sums, again, no luck.
sum(1) = 1
sum(2) = 1 + 3 = 4
sum(3) = 4 + 6 = 10
sum(4) = 10 = 11 = 21
sum(5) = 21 + 31 = 52
Even Wolfram could not find a pattern in your terms
http://www.wolframalpha.com/input/?i=pattern+%7B1,+3,+6,+11,+31,+48,+...%7D
I suspect a typo.
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