Question

if l,m,n are positive integers such that lmn+ln+mn+lm+l+m+n=1000 then l+m+n​

Answer 1

Answer:

e

α→0

lim

(

cosα

n

−1

e

cosα

n

−1

−1

⋅

α

m

cosα

n

−1

)=

2

−e

⇒

α→0

lim

(

α

m

1−cosα

n

)=

2

1

⇒

α→0

lim

(

(α

n

)

2

1−cosα

n

)⋅α

2n−m

=

2

1

⇒2n−m=0

∴

n

m

=2