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Without an equality sign it is no kind of an equation at all.
32
It is not possible to give a conclusive answer because for a recursive relationship of order 1, the first (or 0th) term must be specified.A(n) = (5*n^2 + 3*n + 2*A(1) - 8)/2 for n = 1, 2, 3, ...
Infinitely many. For example: Un+1 = Un + 3 or Un+1 = 2*Un - 1 or Un+1 = 3*Un - 5 or, more generally, Un+1 = k*Un + 7 - 4*k where k is any number. Each one of them will be different from the third term onwards. These are linear patterns. There are quadratic and other recursive relationships.
Three thirds
an = an-1 + d term ar-1 = 11 difference d = -11 ar = ar-1 + d = 11 - 11 = 0 The term 0 follows the term 11.
Without an equality sign it is no kind of an equation at all.
15
It means that each number in the sequence is formed from the previous number by adding d to it. So, 1, 1+d, 1+2d, 1+3d, etc or 4, 4+d, 4+2d, 4+3d, etc
void infinte (long l) { printf ("I will never stop #%ld\n", l); infinte (l+1); }
The sequence is poorly defined. 1+3+5 appears to be a sequence of odd numbers. However, that cannot end in 100: it can attain the values of 99 or 101. Obviously the answer will depend on which one of these is the final number. An alternative is that the sequence is not that of odd numbers but some other sequence: for example, t(n) = (29n3 - 174n2 + 399n - 214)/40 which, for n = 1, 2, 3, generates the sequence 1, 3, 5, 11.35, 26.4, 54.5, 100 whose sum is 201
32
1/3 plus 2/3 plus 3/3 = two
The C++ standard library contains all the pre-defined functions.
It is not possible to give a conclusive answer because for a recursive relationship of order 1, the first (or 0th) term must be specified.A(n) = (5*n^2 + 3*n + 2*A(1) - 8)/2 for n = 1, 2, 3, ...
1 whole, or 1
Yes, this can be done. For example for Fibonacci series. You will find plenty of examples if you google for the types of series you need to be generated.