f(1)=0 and f(m+n)=f(m)+f(n)+4(9mn-1) for natural numbers m and n. What is the value of f(17)?


TCS by SkillRack on 16-Mar-2016 11:37
f(1)=0 and f(m+n)=f(m)+f(n)+4(9mn-1) for natural numbers m and n. What is the value of f(17)?
Siddhant Sharma on 16-Mar-2016 21:08Accepted Solution
f(1) = 0 f(2) = f(1+1) = f(1)+f(1)+4(9×1×1 – 1) = 0+0+4×8 = 32 f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204 f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980 f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260 f(17) = f(1+16) = f(16)+f(1)+4(9×16×1 –1) = 4260+0+ 4×143 = 4832
madhumathi A on 18-Mar-2016 14:34
4832
ANKIT KUMAR CHANDRAKER on 18-Mar-2016 17:18
4832
Solution
f(1) = 0 f(2) = f(1+1) = f(1)+f(1)+4(9×1×1 – 1) = 0+0+4×8 = 32 f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204 f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980 f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260 f(17) = f(1+16) = f(16)+f(1)+4(9×16×1 –1) = 4260+0+ 4×143 = 4832