MBA
Exam
Find the smallest positive integer "n" for which (2^2-1 )(3^2-1)(4^2-1)...(n^2-1) is a Perfect Square? 1) 7 2) 8 3) 9 4) 11
Read Solution (Total 1)
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- nth term = (n^2−1) = (n + 1)(n −1)
⇒series = 1 ×3 × 2 × 4 × 3 × 5 ... × (n − 2) × (n)× (n− 1)× (n + 1) = 2 n (n + 1)× k
because all the other terms are squared.
The fist value of n which 2 n (n + 1) as perfect square is n = 8. - 7 years agoHelpfull: Yes(0) No(1)
MBA Other Question
N is a natural number which gives remainders 1 and 2 when divided by 6 and 5, respectively. All such N’s are written in the ascending order, side by side from left to right. What is the 99th digit from the left? (A) 2(B) 0(C) 1(D) 7 1) A 2) B 3) C 4) D
plz solve this questn and explain it.. i am facing problm in this..a function f(z1,z2,z3........,zn)= f(z1,zn)+ f(z2,z3,z4....., z(n-1))+ (z1+z2+z3.....+zn); for n>0 f(y,z)=f(z,0)+ f(0,y)f(y,0)= y+ f(y-1,0)f(0,y)= y- f(0,y-1)f(0,0)=1find the value of f(1,2,3,4,...........,n), where n is a perfect cube less than 50 and n is greater than 25 1) 1665 2) 1089 3) 729 4) not defined