Given a sequence whose nth term is
T(n) = n2 – (n – 1)2
The task is to evaluate the sum of first n terms i.e.
S(n) = T(1) + T(2) + T(3) + … + T(n)
Print S(n) mod (109 + 7).
Input: n = 3
S(3) = T(1) + T(2) + T(3) = (12 – 02) + (22 – 12) + (32 – 22) = 1 + 3 + 5 = 9
Input: n = 10
Approach: If we try to find out some initial terms of the sequence by putting n = 1, 2, 3, … in T(n) = n2 – (n – 1)2, we find the sequence 1, 3, 5, …
Hence, we find an A.P. where first term is 1 and d (common difference between consecutive
terms) is 2.
The formula for the sum of n terms of A.P is
S(n) = n / 2 [ 2 * a + (n – 1) * d ]
where a is the first term.
So, putting a = 1 and d = 2, we get
S(n) = n2
Below is the implementation of above approach:
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- Divisibility by 12 for a large number
- Divisible by 37 for large numbers
- Recursive sum of digit in n^x, where n and x are very large
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