1ohh! great vivek !
find sum to n terms of the series
0, 24, 120, 360, 840, 1680 ..........
find sum in term of n only
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22 Answers
1The series is 4!(3c4+4c4 + 5c4 + 6c4 + ....... nc4 + n+1c4 + n+2c4 )
but 3c4=0
finding 4c4 + 5c4 + 6c4 + ....... nc4 + n+1c4 + n+2c4 is equal to the coeff of x4 in S,
where S = (1+x)4+(1+x)5 + ... (1+x)n+2 = ((1+x)n-1-1)(1+x)4/(x)
so we have tofind coeff of x in (1+x)n+3/x which is n+3C5
1we must refer it to nishant otherwise we have to calculate the summation of n4
1no vivek i was not referring to this point ..
and was trying to calculate the sum via ncr method but not been able to solve it
the series will be 4
1vivek u got the logic ...just a calculation mistake ...i think can u again check ur ans
1if 0 is the first term , then the sum should be 4!n+3c5
the first term is 3c4 = 0
1the sum is = 4!/0! + 5!/1! +6!/2! + 7!/3! + ................
= 4! (4c4 + 5c4 + ..... n+3c4)
=4! (n+4c5 )
1t1= 0 | = 120 t2=120 | = 240 t3=360 |= 480 t4 = 840 | =840 t5= 1680 | = next diff 1680 .. like this ... only possible if 24 is not a term... if it is a trem,,, then i am sorry for this post... i will think again...
1is this series correct ?? pls confirm...
i mean i am having a feeling like 24 will not be there...