Solve It.......2

$2^2\times(2+1)=12$

$3^2\times(3+1)=36$

$\rightarrow 4^2\times(4+1)=80$

(f(n) = (n + 1) n^2, so f(4) = 5 16 = 80. Ed Gray

$2^2 \times (2+1) = 12$ $3^3 \times (3+1) = 36$ $4^4 \times (4+1) = \boxed{80}$ $5^5 \times (5+1) = 150$ $6^6 \times (6+1) = 36 \times 7 = 252$ $7^7 \times (7+1) = 49 \times 8 = 372$

2 X 6=12 3X12=36 4X20=80 5X30=150 6X42=252 7X56=392 We can see that there is a pattern with the nos. that are getting multiplied with the nos.(1,2,3,...) each time..that are (6,8,10,12,14) gettin' added with the previous multiplier...hope i'm correct..thanks.