Find the summation,
9+16+29+54+103+............nth term
Hmm. Using Newton's Divided Differences, we get
Tn = 1/4 (x^4-6x^3+23x^2-14x+32)
Sn = 1/20 x(x^4-5x^3+25x^2+15x+144)
Seems a bit complicated for the average induction problem.
To find the summation of the given series, we observe a pattern in the terms:
9, 16, 29, 54, 103, ...
Looking at the differences between consecutive terms, we can see that they form a sequence like this:
7, 13, 25, 49, ...
We can notice that each term in this sequence can be written as the result of adding a power of two (starting from 1) to a constant value. In other words, the nth term of this sequence can be written as:
2^n - 1
Using this pattern, we can write the nth term of the original series as:
T(n) = (2^n - 1)^2
Now, let's express the summation of the series up to the nth term. We will represent the sum as S(n):
S(n) = 9 + 16 + 29 + 54 + 103 + ... + T(n)
To find a general formula for S(n), we need to express each term in the series in terms of n:
S(n) = (2^2 - 1)^2 + (2^3 - 1)^2 + (2^4 - 1)^2 + ... + (2^n - 1)^2
Expanding this out, we get:
S(n) = (4 - 2 + 1)^2 + (8 - 3)^2 + (16 - 4)^2 + ... + (2^n - 1)^2
Simplifying further, we have:
S(n) = (3^2)^2 + (5^2)^2 + (7^2)^2 + ... + (2^n - 1)^2
Now, by using the sum of squares formula, we can find a simplified expression for S(n):
S(n) = (1/6) * [n * (n + 1) * (2n + 1)] - (n * (n + 1) / 2) + n
Therefore, the general formula for the summation of the series up to the nth term is:
S(n) = (1/6) * [n * (n + 1) * (2n + 1)] - (n * (n + 1) / 2) + n