I need to prove by induction that:

1/(1x3) + 1/(2x4) +... + 1/n(n+2) = n(3n+5)/4(n+1)(n+2)

Thanks

1/(1x3) + 1/(2x4) +... + 1/n(n+2) = n(3n+5)/4(n+1)(n+2)


n=1) 1/(1.3) = 1/3
1.(3.1+5)/4(1+1)(1+2) = 8/24 = 1/3

n-1=> n)

1/(1x3) + 1/(2x4) +... + 1/(n-1)(n+1) = (n-1)(3n+2)/4n(n+1)

adding 1/n(n+2)

1/(1x3) + 1/(2x4) +... +1/n(n+2) = (n-1)(3n+2)/4n(n+1) + 1/n(n+2)
= ( (n-1)(3n+2)(n+2)+ 4(n+1) ) / (4n(n+1)(n+2))
= (3n^3 + 5n^2) / (4n(n+1)(n+2))
= n(3n+5)/4(n+1)(n+2)

1/(1x3) + 1/(2x4) +... + 1/n(n+2) = n(3n+5)/4(n+1)(n+2)

True for n=1

1/(1*3) = 1(8)/[4(2)(3)]

since both sides reduce to 1/3

Assume that

1/(1x3) + 1/(2x4) +... + 1/k(k+2) = k(3k+5)/4(k+1)(k+2)

Add 1/(k+1)(k+3) to both sides

Then

1/(1x3) + 1/(2x4) +... +1/(k+1)(k+3) = k(3k+5)/4(k+1)(k+2) +1/(k+1)(k+3)

= [k(3k+5)(k+3)+4(k+2)]/[4(k+1)(k+2)(k+3)]

= [k(3k^2 +14k+15) +4k+8]/[4(k+1)(k+2)(k+3)]

= [3k^3 + 14k^2 +19k +8]/[4(k+1)(k+2)(k+3)]

= [(k+1)(3k^2+11k+8)]/[4(k+1)(k+2)(k+3)]

= (3k^2 +11k + 8)/[4(k+2)(k+3)]

= (3k+8)(k+1)/[4(k+2)(k+3)]

= [(k+1)[3(k+1) + 5]]/[4(k+1+1)(k+1+2)]

So from assuming true for n=k we showed that true for n=k+1

Since true for n=1,true for n=2,3,.
By Math Induction true for all positive integers

Thanks! :D