Question

1. 3. 5
- + - + -
3. 4. 6​

Answer 1

Answer:

1/2

Step-by-step explanation:

as we know, (1+x)n=1+n(n−1)x2+n(n−1)(n−2)x23⋅2⋅1+......

(1+x)−12=1+(−12)(−32)x2+(−12)(−32)(−52)x23⋅2⋅1+.....

(1+x)−12=1+1⋅3⋅x2⋅4−1⋅3⋅5x22⋅4⋅6+...

12(1−x)−12=1−x2+1⋅3⋅x22⋅4+1⋅3⋅5x32⋅4⋅6+....

=12−x2+1⋅3⋅x22⋅4⋅2+1⋅3⋅52⋅4⋅6⋅2x3+.....

integrating both sides

12∫10(1−x)−12=12−[x22⋅4]+1⋅32⋅4⋅2[x23]+1⋅3⋅52⋅4⋅6⋅2[x44]

12∫10(1−x)−12=12−[x22⋅4]

12∫10(1−x)−12=12+[12⋅4+1⋅32⋅4⋅6+1⋅3⋅52⋅4⋅6⋅8+....]

S=12∫10(1−x)9−12)−12)=12⋅4+1⋅32⋅4⋅6+1⋅3⋅52⋅4⋅6⋅8+...

=12[∫01−2tdtt−1]

=12[−2[t]−1=12[2−1]

=12