Question

f(x)=4x-5 f(1)+f(2)+f(3)+....+f(10)?

Answer 1

f(1)=4·1-5⇔f(1)=-1
f(2)=4·2-5⇔f(2)=3
f(3)=4·3-5⇔f(3)=7
f(4)=4·4-5⇔f(4)=11
f(5)=4·5-5⇔f(5)=15
f(6)=4·6-5⇔f(6)=19
f(7)=4·7-5⇔f(7)=23
f(8)=4·8-5⇔f(8)=27
f(9)=4·9-5⇔f(9)=31
f(10)=4·10-5⇔f(10)=55


S=(-1+55)*10/2
S=54*10/2
S=540/2
S=270