1i totally agree whd nishant.........
Find the coefficient of x50 in
(1+x)+2(1+x)2+3(1+x)3+....+1000(1+x)1000
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7 Answers
62Hint.
take 1+x=y
then find the sum.
put back y
then find coeff of x50
can u try this way?
1hmmm....
-(ln(1-x)=x+x^2/2+x^3/3.......
diff it we get
1/(1-x)=1+x+x^2+x^3..........
diff again we get
(-1)/(1-x)^2=1+2x+3x^2+4x^3.......
nw multiplyin by x we get
(-x)/(1-x)=x+2x^2+3x^3.......
just substitute x=x+1
u get ur question ........ie
(1-)(x+1)/(-x)=(1+x)+2(1+x)^2+3(1+x)^3............
so u basically need 2 find d ter x^50 in (1+1/x)........
wich u can find i guess
(m bad at that part :()
1hey here your exp. goes up to infinity!!
but in the question it is upto 1000
you cant do that!!
1There is no term of x50 until 50(1+x)50
So the coefficient of x50 is derived from........
50(1+x)50+51(1+x)51+52(1+x)52+....+1000(1+x)1000, whose value is...
=5050C50+5151C50+5252C50+.....+10001000C50
I am not very good at Binomial, so I dont know if Anything can be done beyond what I got[3]
Someone please try [5]
62y=1+x
the given sum is Σn.yn
\large \frac{y}{(1-y)^2}(1000y^{1001}-1001y^{1000}+1)
put back y=1+x
\large \frac{x+1}{(x)^2}(1000(x+1)^{1001}-1001(x+1)^{1000}+1)
this is same as finding the coeff of x50 in the above expression.
which is same as finding the coeff of x^52 in \large \frac{x+1}{1}(1000(x+1)^{1001}-1001(x+1)^{1000}+1)
I hope this helps now :)