2011/004) Using mathematical induction prove that tan ^-1 (1/3) + tan ^-1 (1/7) + … tan ^-1 (1/(n^2+n+ 1) = tan ^-1(n/(n+2))

in mathematical induction there are 2 steps

number one base step

for n =1 LHS = tan ^-1 (1/3) and RHS = tan ^-1 (1/(1+2)) = tan ^-1 (1/3)

hence true

then the induction step

let it be true for n = k that is

tan ^-1 (1/3) + tan ^-1 (1/7) + … tan ^-1 (1/(k^2+k+ 1) = tan ^-1(k/(k+2))

for n = k + 1 we have
tan ^-1 (1/3) + tan ^-1 (1/7) + … tan ^-1 (1/(k^2+k+ 1) + tan ^-1(1/(k+1)^2 + k+ 1+ 1)
= tan ^-1(1/(k+2)) + + tan ^-1(1/(k+1)^2 + k+ 1+ 1)
= tan ^-1(1/(k+2)) + + tan ^-1(1/(k^2+3k + 3)

The RHS = tan ^-1((k+1)/(k+3))


As tan (A+ B) = (tan A + tan B)/(1- tan A tan B)

So A + B = tan ^-1(tan A + tan B)/(1- tan A tan B))

So tan ^-1(k/(k+2)) + tan ^-1(1/(k^2+3k + 3)
= tan ^-1 ( k/(k+2) + 1/(k^2+3k + 3)/(1-k/(k+2)1/(k^3+3k+ 3)
= tan ^-1(( k^3+3k^2+3k + k+ 2)/((k+2)(k^2+3k+3)-k)
= tan ^-1((k^3+3k^2 +4k + 2) /(k^3+ 5k^2 + 8 k + 6))

We have k^3+3k^2+ 4k + 2 = (k^3+ 3k^2 + 3 k + 1) + (k+1) = (k+1)^3 + k+ 1= (k+1)(k^2 + 2k + 2)

(k^3+ 5k^2 + 8 k + 6) = k^3 + 3k^2 + 2k^2 + 8k + 6
= k^2(k+3) + 2( k^2 + 4k + 3) = k^2(k+3) + 2( k+1) (k+ 3) = (k+ 3)( k^2 + 2k + 2)

So tan ^-1((k^3+3k^2 +4k + 2) /(k^3+ 5k^2 + 8 k + 6)) = tan ^-1 ((k+1)/(k+ 3)

= RHS

So induction step is proved

hence proved