A = 2•3 + 3•4 + 4•5 + ... + 11 • 12 If the second factor of each term in the sum is increased by 1, how much does the number A increase? A) 63 B) 64 C) 65 D) 66 E) 68

Let's denote the sum as S .
S = 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + \cdots + 11 \cdot 12
To simplify this expression, we can factor out common terms:
S = 2(3) + 3(4) + 4(5) + \cdots + 11(12)
S=(3^2-3)+(4^2-4)+\cdots+(11^2-11)+\left(12^2-12\right)
S=3^2+4^2+\cdots+11^2+12^2-(3+4+\cdots+11+12)
S=(3^2+4^2+\cdots+11^2+12^2)-\left(\frac{13\cdot12}{2}\right)
S=\frac{12\cdot13\cdot25}{6}-5-6\cdot13

Now, if the second factor of each term is increased by 1, we get:
S' = 2(3+1) + 3(4+1) + 4(5+1) + \cdots + 11(12+1)
= S + 2+3+........11
Using the same approach as before, we can find the new sum S' :
= S+ 12*11/2 -1 = S+ 65




The increase in the sum is S^{\prime}-S=65 ,