1057wat happened?no one cud find out a answer?
find the sum to infinity....1/2.4+1.3/2.4.6+1.3.5/2.4.6.8 .......
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9 Answers
1057well....i tried it......found out the nth term...
Tn=1.3.5........upto n/2 terms2(1.2.3.4......n/2 terms
now remove the 1.3.5.7........n/2 from both the numerator and denominator.....
now Tn=12*(n/2)!
but now after that????
1057sorry dis abve post is wrong.....
Tn=1.3.5........upto n/2 terms
2(1.2.3.4......n/2 terms
now remove the 1.3.5.7........n/2 from both the numerator and denominator.....
now Tn=1122*(n/4)!
21This has got nothing to do with progression..try binomial theorem for fractional index..( i hope this will work)
1From Shubodip's Hint :
\left(1-x \right)^{-\frac{1}{2}}=1+\frac{x}{2}+\frac{1.3}{2^2 2!}x^2+\frac{1.3.5}{2^3 3!}x^3+\cdots \\ \\ \frac{1}{2}\int_{0}^{1}{\left(1-x \right)^{-\frac{1}{2}}}\mathrm{dx} =\frac{1}{2}+\frac{1^2}{2^3}+\frac{1.3}{2^3 3!}+\frac{1.3.5}{2^4 4!}+\cdots \infty \\ \\ S=\boxed{\left( \frac{1}{2}\int_0^{1}\left(1-x \right)^{-\frac{1}{2}}\right)-1} \\ \\ \boxed{S=\frac{1}{2}}
7Wow i can do this now......
S=1/2.4 +............
or 2S= 1/4 +1.3/4.6 +1.3.5/4.6.8 ......................
= 1-3.4 + 1.3(1-5.6)/4 + 1.3.5(1-7/8)?4.6. .............
= 1- 3/4 +3/4 -.............................
=1
SO S= 1/2...... (But ther is 1 flaw which i have delberately overlooked...... .......But the ans is correct...At last i did it....