EAMCET Engineering 2008 Solved Paper

Given that, α=‌

5

2!3

+‌

5⋅7

3!32

+‌

5⋅7⋅9

4!33

+... . . . (i)
We know that,
(1+x)n=‌1+‌

nx

1!

+‌

n(n−1)

2!

x2
‌+‌

n(n−1)(n−2)

3!

x3+... . . . (ii)
On comparing Eqs. (i) and (ii), with respect to factorial
n(n−1)x2‌=‌

5

3

. . . (iii)
n(n−1)(n−2)x3‌=‌

5⋅7

32

. . . (iv)
and n(n−1)(n−2)(n−3)x4=‌

5⋅7⋅9

33

. . . (v)
On dividing Eq. (iv) by (iii) and Eq. (v) by (iv), we get
(n−2)x=‌

7

3

. . . (vi)
and
(n−3)x=3 . . . (vii)
Again, dividing Eq. (vi) by (vii), we get
‌‌

n−2

n−3

=‌

7

9

‌⇒‌‌9n−18=7n−21
‌⇒‌‌2n=−3
‌⇒‌‌n=−‌

3

2

On putting the value of n in Eq. (vi), we get
(−‌

3

2

−2)x=‌

7

3

⇒x=−‌

2

3

∴ From Eq. (ii),
(1−‌

2

3

)−3∕2‌=1+1+‌

5

2!3

+‌

5⋅7

3!32

+...
⇒‌‌33∕2−2‌=‌

5

2!3

+‌

5⋅7

3!32

+...
⇒‌‌α=33∕2−2‌‌[‌ from Eq. (i) ‌]
‌ Now, ‌α2+4α‌=(33∕2−2)2+4(33∕2−2)
=‌27+4−4⋅33∕2+4⋅33∕2−8
=‌23