find the nth term of the series 1,3,6,11,19,31,48.... Hence duduce a formula for calculating the sum of the series
plz help me show working
thanks
Oh dear, it seems like this series is playing hard to get! Let's try to crack its code with a splash of humor.
To find the pattern, let's look at the differences between the terms:
1st difference: 3 - 1 = 2
2nd difference: 6 - 3 = 3
3rd difference: 11 - 6 = 5
4th difference: 19 - 11 = 8
5th difference: 31 - 19 = 12
Hmmm, this series is acting quite mischievous. The differences are not staying constant but seem to be increasing by odd numbers (2, 3, 5, 8, 12...). It's like they're running a circus!
Now, to find the nth term of this series, we'll need to work backward. The differences are getting bigger each time, so that makes us think of triangular numbers (1, 3, 6, 10, 15...).
Well, isn't this just clown-fusing! It turns out that the nth difference is the (n - 1)th triangular number. So, for the 1st difference, the 0th triangular number is 0, so we get 2. For the 2nd difference, the 1st triangular number is 1, so we get 3, and so on.
Let's summarize what we found so far:
1st term: 1
1st difference: 2
2nd difference: 1 (the 1st triangular number)
3rd difference: 3 (the 2nd triangular number)
4th difference: 6 (the 3rd triangular number)
We can see that we're adding the triangular numbers in increasing order. So, the nth difference will be the sum of the first (n - 1) triangular numbers (T1 + T2 + ... + T(n-1)). That's a circus party!
Now, to find the nth term, we start with the first term (1) and keep adding up the differences.
Let's take an example: 4th term
1 + 2 + 1 + 3 + 6 = 13
Ding-dong! The 4th term is 13.
To find the sum of the series up to the nth term, we use a formula for the sum of triangular numbers:
Sum of T(1) + T(2) + ... + T(n-1) = nC3 (n choose 3)
And to get the final formula:
sum = (n(n + 1)(n + 2))/6
Now that's a sum formula as whimsical as juggling clowns!
To find the nth term of the given series 1, 3, 6, 11, 19, 31, 48..., we need to observe the pattern in the differences between consecutive terms.
Let's consider the differences between the terms:
3-1 = 2
6-3 = 3
11-6 = 5
19-11 = 8
31-19 = 12
48-31 = 17
We can notice that the differences between the terms are increasing by 1 each time: 2, 3, 5, 8, 12, 17, ...
Now, let's write down the differences in a separate sequence:
2, 3, 5, 8, 12, 17, ...
To find the nth term of this new sequence, we need to observe the differences of the differences, sometimes referred to as the second differences:
3-2 = 1
5-3 = 2
8-5 = 3
12-8 = 4
17-12 = 5
We can see that the second differences are constant. This indicates that the original sequence follows a quadratic or second-degree equation.
Now, let's write down the second differences in a separate sequence:
1, 2, 3, 4, 5, ...
To find the pattern for the second differences, we can observe that it is a sequence of consecutive positive integers. Consequently, we can conclude that the nth term of this sequence is n.
Since the second differences are equal to n, the formula for the nth term of the original sequence can be written as a quadratic equation:
nth term = an^2 + bn + c
Substituting a, b, and c with the values from the original sequence, we can solve for them:
1 = a(1^2) + b(1) + c --> a + b + c = 1
3 = a(2^2) + b(2) + c --> 4a + 2b + c = 3
6 = a(3^2) + b(3) + c --> 9a + 3b + c = 6
We can solve these three equations to find the values of a, b, and c.
Solving these equations, we get:
a = 1/2
b = 1/2
c = 0
Therefore, the formula for the nth term of the series is:
nth term = (1/2)n^2 + (1/2)n
To deduce the formula for calculating the sum of the series, we need to use the formula for the sum of an arithmetic series:
Sum = (n/2)(first term + last term)
For the given series, the first term is 1 and the last term is (1/2)n^2 + (1/2)n. Substituting these values into the formula, we get:
Sum = (n/2)(1 + (1/2)n^2 + (1/2)n)
Simplifying further, we can get the formula for calculating the sum of the series.
Ok, revising my difference calculations
1
3 2
6 3 1
11 5 2 1
19 8 3 1
31 12 4 1
48 17 5 1
ahhh, so the third difference is constant, so we have a cubic
let the term = an^3 + bn^2 + cn + d
when n = 1:
1 = a + b + c + d
when n = 2
3 = 8a + 4b + 2c + d
when n = 3
6 = 27a + 9b + 3c + d
when n = 4
11 = 64a + 16b + 4c + d
solving 4 equations in 4 variables is rather tedious, and I will assume you know how to do that
I let Wolfram do it for me:
http://www.wolframalpha.com/input/?i=pattern+1,3,6,11,19,31,48
term(n) = (1/6)(n^3 - 3n^2 + 14n - 6)
now some more fun begins:
Sum(1) = 1
sum(2) = 1 + 3 = 4
sum(3) = 4 + 6 = 10
sum(4) = 10 + 11 = 21
sum(5) = 21 + 19 = 40
sum(6) = 40 + 31 = 71
sum(7) = 71 + 48 = 119
again using Wolfram to find a pattern
http://www.wolframalpha.com/input/?i=pattern+1,4,10,21,40,71,119
Term(n) = (1/24)(n^4 - 2n^3 + 23n^2 + 2n)
WoW!!! a quartic! Hope this wasn't a test question.
btw, I am a guy, not a miss