Question

(i). F(1) = 1
(ii). F(1) + 2F(2) + 3F(3) + ... + nF(n) = n(n + 1)F(n)
for n > 2 calculate the value of F(2015)

Answer 1

2015(2015+1)F(2015)

Answer 2

that's just the RHS, what about with the LHS ?

Answer 3

@rational

Answer 4

F(1) + 2F(2) + 3F(3) + ... + 2015F(2015) = 2015(2015 + 1)F(2015)

Answer 5

we have
\[ F(1) + 2F(2) + 3F(3) + \cdots + (n-1)F(n-1) + \color{blue}{nF(n)} = n(n + 1)F(n)\]

send that blue term to other side and get

\[ F(1) + 2F(2) + 3F(3) + \cdots + (n-1)F(n-1) = [n(n + 1) - n]F(n)\]

\[ F(1) + 2F(2) + 3F(3) + \cdots + (n-1)F(n-1) = n^2F(n)\]

yes ?

Answer 6

Next observe that left hand side becomes \(n(n-1)F(n-1)\) and you get a recurrence relation :

\[n(n-1)F(n-1) = n^2F(n)\]

Answer 7

cancel \(n\) both sides and use induction to conclude
\[F(n) = \dfrac{1}{n}\]

Answer 8

for n = 2014
F(1)+2F(2)+3F(3)+...+2014F(2014)=2014(2014+1)F(2014)=2014(2015)f(2014)
for n = 2015
F(1)+2F(2)+3F(3)+...+2015F(2015)=2015(2015+1)F(2015)=2015(2016)f(2015)
substract both equation gives us :
2015f(2015) = 2015(2016)f(2015) - 2014(2015)f(2014)
still there is the f(2014) --"

Answer 9

ok, im following till this part
n(n - 1)F(n-1) = n^2F(n)
(n - 1)F(n-1) = n F(n)
but how you establish that F(n) = 1/n ?

Answer 10

(n - 1)F(n-1) = n F(n)
F(n)/F(n - 1) = (n - 1)/n
F(n)/F(n - 1) = 1 - 1/n
seems the telescopic then, right ?

Answer 11

F(n)/F(n - 1) = 1 - 1/n

n=2
F(2)/F(1) = 1 - 1/2 = 1/2

n=3
F(3)/F(2) = 1 - 1/3 = 2/3

n=4
F(4)/F(3) = 1 - 1/4 = 3/4
.
.

n=2014
F(2014)/F(2013) = 1 - 1/2014 = 2013/2014

n=2015
F(2015)/F(2014) = 1 - 1/2015 = 2014/2015

multiply all get
F(2)/F(1) x F(3)/F(2) x F(4)/F(3) ... F(2014)/F(2013) x F(2015)/F(2014 = 1/2 x 2/3 x 3/4 x ... x 2013/2014 x 2014/2015

F(2015)/F(1) = 1/2015
F(2015) = 1/2015
hehe