39may i know from where the q has come??
plz solve this one :
[1]
q2
try this one also its not my doubt but wanted to share
if
log(\frac{1}{1+x+x^2 +x^3 })
is expanded on the ascendind powers of x then prove that
coefficient of
xn in the expansion is
\frac{-1}{n}\ \ \ \ \ \ \ \ \ \ if\ n \ is \ odd\ or \ of \ form\ 4m+2 \\ \\ and \\ \\ \frac{3}{n}\ \ \ \ \ \ \ \ \ \ \ if \ n \ is \ of \ form%20%5C%204m
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8 Answers
106Q2.
log(11+x+x2+x3) = log(1(1+x2)(1+x))
= -(log(1+x) + log(1+x2)
= -(x-x2/2+x3/3-... + x2-x4/2+x6/3-...)
So coeff of xn (if odd) is -1/n
Now if of the form 4m, then -(-1/4m - 1/2m) = 3/4m = 3/n
If of the form 4m+2 then -(-1/(4m+2) + 1/(2m+1)) = -1/(4m+2) = -1/n
66The given sum
S=1+\dfrac{1}{3}+\dfrac{1\cdot 3}{3\cdot6}+\dfrac{1\cdot 3\cdot 5}{3\cdot6\cdot 9}+\ldots
=1+\dfrac{1}{3}+\dfrac{1\cdot 3}{3^2\cdot2!}+\dfrac{1\cdot 3\cdot 5}{3^3\cdot 3!}+\ldots
=1-\dfrac{1}{2}\left(\dfrac{-2}{3}\right)+\dfrac{\left(\frac{-1}{2}\right)\left(\frac{-1}{2}-1\right)}{2!}\left(\dfrac{-2}{3}\right)^2+ \dfrac{\left(\frac{-1}{2}\right)\left(\frac{-1}{2}-1\right)\left(\frac{-1}{2}-2\right)}{3!}\left(\dfrac{-2}{3}\right)^3+\ldots
=\left(1-\dfrac{2}{3}\right)^{-1/2}
=\sqrt{3}
24upto 2nd step its OK...
but sir plz tell me what was inspiration for 3rd step ??
24that is obvious....but how did that idea strike ???
and plz not the same answer this time too..PRACTICE [3]
No offence intended[1]
