Mathematics Times – July 2019

2.Sol: Given that

3 2 4

1

( )

n

k

ak bk ck d n



    

3 2 4

1 1 1 1

n n n n

k k k k

a k b k c k d n

   

      

2

( 1) ( 1)(2 1)

2 6

n n n n n

a b

      

   

   

( 1) 4

2

n n

c dn n

  

   

 

n^4 (12 3 ) (4 6 ) (6 6 3 ) a n b a n c b a^3  ^2  

  n c b(6 2 12 ) 0d 

equating corresponding coefficients, we get

12 3 0;4 6 0   a b a ;

(^) 6 6 3 0;6 2 12 0c b a     c b d
which yields a b   4, 6, 4c and d  1.
Therefore a b c d    15.
3.Sol: nth term of the given sequence is
(^2 1)!
T r rr   r
(^)    ( 1)! ( 2) !r^2 r r r
    ( 1)( 1)! ( 2) !r r r r
Therefore the sum is

 Tr ( 1)( 1)! ( 2) !r   r r r

           3! 2 4! 3 5! ... ( 1)( 1)! 1 1 3!n n

     2 4! 3 5! ( 2) !n n

   1 ( 1)( 1)!n n

4.Sol: 10
2

x y

a b



  for some 1  a 9 ,

0  b 9.

xy  10 b a

2 2

100 20^222

4

x xy y

a ab b

 

  

xy100 20b^2  ab a^2

2 2 2 2 2

2 2

4 4 2

x xy y x xy y x y

xy

      

   

 

99 99 99(a^2  b^2  a b^2 ^2 )

(^) x y 2 99(a b^2  ^2 ) 6 11(a b^2 ^2 )
Note that in order for x y to be integer,,
(a b^2 ^2 )has to be 111n for some perfect square
n. Since a is atmost 9,n 1 or 4
If n  1,x y 66 then 11 n with
n 1 may be obtained with a 6 and b 5
as a b^2    ^2 36 25 11.
Therefore x y 2(10(6) 5) 130  , if
n  4,x y 132. We need to show that
a b^2  ^244 is impossible.
( )( ) 44a b a b     a b 1 , or 2 or 4
and a b 44, 22, 11 respectively. And since
a b 18,a b  11 , a b  4 , but there is
no integer solution for a, b.
Note: it is easy to see that ( , ) (6,5)a b  is
the only solution. This yields ( , ) (98,32)x y .
Their arithmetic mean is 65 and their
geometric mean is 56.
5.Sol: Let a ar ar, ,^2 are sides of a triangle. The
longest side can’t be longer than the sum of
the other two sides. That is
a ar ar ^2
^ a r ar(1 ) ^2
^1  r r^2
^ r r^2   1 0
1 5
2
r  for an increasing progression and
2.Sol:
3.Sol:
4.Sol:
5.Sol:
6.Sol:
1 5
2
r
 
 for a decreasing progression.
6.Sol: Given recursive sequence an  3 an 1 1 is
written as an3(3 1) 1 3an 2   ^2 an 2  3 1,
likewise more generally an   3 na 0 3 ...n^2
 3 1. Therefore
2010
0
3 1
3
2
n
an a

  , but