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To CREATE AND PRINT the SERIES 1,2,2,3,3,3,,4,4,4,4…n,n,n,n,n.

Submitted on: 11/7/2015 1:42:54 AM
Level: Intermediate
User Rating: Unrated
Compatibility: C, C++ (general)
Views: 3623
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     In the series ......1,2,2,3,3,3,4,4,4,4,...............n,n,n,n. 1) number of Terms up to last term which is n is =n(n+1)/2 2) Sum Of Terms =1+2+2+3+3+3+4+4+4+4+..............+n+n+n=1+2*2+3*3+4*4+5*5+....n*n =n(n+1)(6n+1)/2 3) Find The General Term..... Here t1=1 ; t2 to t3 =2 ; t4 to t6=3 etc.... Total Number of Terms =n(n+1)/2 n(n+1)/2 th term is n and (n-1)n/2 th term is n-1 which is the last term which is n-1 Then [ (n-1)n/2 + 1 ] th term is n since it is immediately after the last term which is n-1 ok then in general [ (n-1)*n/2 +1 ] th term to [ n(n+1)/2] th term =n where n=1,2,3,4,5....... is the general term ok .You please check it by putting n=1,2,3,4,5..... Now can you find the 5000 th term .Please find it ? Put n(n+1)/2 = 5000 and solve for n ; we will get n^2 +n-10000=0 n= [ -1+Sqroot(1+40001) ] / 2 ; n= 99.50124999==100 ok put n=100 in [ (n-1)*n / 2 + 1 ] & [ n*(n+1) /2 ] we will get 4951 to 5050 ok then 5000th term is 100 ..Is it right ….? In Sequel with the Problem and solution of the Series 1,2,2,3,3,3,,4,4,4,4…n,n,n,n,n. I wrote a Code to print this series up to the last term which is n.First we will input Number of Terms in to terms var. Then the Equation n*(n+1)/2=terms is solved and foind the n where n will be the last term of series .We will calculate (n-1)*n/2 + 1 & n*(n+1)/2 and test whether terms var is between these two values .If yes We will store the series up to the term n*(n+1)/2 and in to an Array and it is printed .I am attaching the Code here .
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