Suppose we want to prove the statement S(n): "If n â¥ 2, the sum of the integers 2 through n is (n+2)(n-1)/2" by induction on n. To prove the inductive step, we can make use of the fact that 2+3+4+...+(n+1) = (2+3+4+...+n) + (n+1) Find, in the list below an equality that we may prove to conclude the inductive part.

a. If n â¥ 3 then (n+2)(n-1)/2 + n + 1 = (n+3)(n)/2
b. If n â¥ 1 then (n+2)(n-1)/2 + n + 1 = (n+3)(n)/2
c. If n â¥ 2 then (n+2)(n-1)/2 + n + 1 = (n+3)(n)/2
d. If n â¥ 1 then (n+2)(n-1)/2 + n + 1 = n(n+3)/2

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codiepienagoya codiepienagoya · Answer 1 · 2021-02-12T07:34:43+01:00

Answer:

The answer is "Choice c".

Explanation:

Please find the complete question in the attached file.

To begin with, allow its principle of the numerically solving to be recognized, three stages are concerned.

1. Topic n=1

2. Suppose n to be true

3. Display n+1 it retains

We have LHS as 2+3+ for the third step now [tex](n+1) = (2+3+.. \& n) + n+1[/tex]

We can now replace the bracket of RHS by [tex]\frac{(n+2)(n-1)}{2},[/tex]as we assumed its valid for n in step 2

if we do that we get

[tex]= \frac{(n+2)(n-1)}{2+(n+1)}\\\\= \frac{(n^2-n+2n-2+2n+2)}{2}\\\\= \frac{(n^2+3n)}{2}\\\\= \frac{n(n+3)}{2}[/tex]