39i do not think the same trick always applies...
remember having tried it but being unsuccesful o get any general method..
while solving this sum:
find
C0+C1/2 +C2 /3 +.............+Cn/n+1
there is an alternate method given in BT
that is
multiply and divide by( n+1)
\frac{1}{n+1}[(n+1)(1+\frac{n}{2} + \frac{n(n-1)}{1.2.3}+...........]
then the expresion reduces to
\frac{1}{n+1}[^{n+1} C_{1}+^{n+1}C_{2}+.............]
= \frac{1}{n+1}[2^{n+1}-1]
but i am unable to apply this trick in other problems
can anyone please discuss some more examples on solving some through this technique..............
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7 Answers
39
1okay,one more example werin the method is used
2.nC0+22 nC1/2 +.........2n+1 nCn/n+1
again he has multiplied and divided by n+1,and has done that sum....
i think this trick can be done only when denominator is n+1....
can u plz tell me why are putting the limits as 0 to 1 while doing the sum via integration
i mean what is the concept behind the limits
1I think we integrate withiin the limits 0 to 1 so as to cancel out x in the expansions which is only possible for x=1 or x=0.
1@soumy , can u elaborate more on what u r saying .if possible can u give an example an explain
1Take for example we want to find c01 + c12 etc.
we take (1+x)n = c0 + c1x + c2x2 + .......
now we integrate both the sides within limits 0 to 1
∫(1+x)ndx = ∫(c0 + c1x + c2x2 ....)dx
so we get [ (1+x)n+1/n+1]01 = (c0x + c1 x 2/2+ c2x3/3 )01 we do this because we want to cancel out x-s from the binomial co-effs which is possible for x=1 or 0 otherwise again some numbers will be multiplied with them.
so we get 2n+1-1n+1 = c01 + c1/2 + c3/3 etc which we wnated.