wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let α and β be the roots of x2x1=0,with α > β, For all positive integers n, define,
an=αnβnαβ,n1
b1=1 and bn=an1+an+1,n2.
Then which of the following option is/are correct?

A
a1+a2+a3+......+an=an+21 such that n1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
n=1an10n=1089
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
n=1bn10n=889
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
bn=αn+βn such that n1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D bn=αn+βn such that n1
α,β are roots of equation x2x1=0,
So,
α+β=1, αβ=1,
α,β=+1±52, αβ=5
α2=(1+52)2=3+52
β2=(152)2=352
L.H.S =a1+a2+a3+......+an
=(α+α2+α3+...+αn)(β+β2+β3+...+βn)αβ
=α(1αn)(αβ)(1α)β(1βn)(αβ)(1β)
α2α1=0α21=αα+1=α1α
=((α+1)(1αn))+(1+β)(1βn)αβ
=α2(1αn)+β2(1βn)αβ=(αn+2βn+2)+(β2α2)αβ
=αn+2βn+2αβα2β2αβ=an+2(α+β)=an+21

n=1an10n=αnβn(αβ)10n=(α10)n(β10)nαβ
n=1(α10)n=α101α10=α10α
Similarly,
n=1(β10)n=β10β
n=1an10n=α10αβ10βαβ=α(10β)β(10α)(αβ)(10α)(10β)
=10α10β(αβ)(10α)(10β)=10(10α)(10β)
=10[101+52][10152]
=10×4(2015)(201+5)=40192(5)2=40356=1089
bn=an1+an+1
=αn1βn1αβ+αn+1βn+1αβ
=αn1+αn+1αββn+1+βn1αβ=αn(α2+1α)βn(β2+1β)5
=αnβ(α2+1)βnα(β2+1)αβ5=αn152(1+3+52)βn1+52(1+352)1×5
=αn(15)(5+5)βn(1+5)(55)4×5
=αn(45)βn(45)45=αn+βn
n=1bn10n=αn+βn10n=(α10)n+(β10)n
n=1(α10)n=α101α10=α10α
Similarly, n=1(β10)n=β10β
n=1bn10n=α10α+β10β=α(10β)+β(10α)(10α)(10β)

=10(α+β)2αβ(10α)(10β)=(10×1)2(1)[101+52][10152]
=(10+2)×4[195][19+5]=12×4(3615)=48356=1289

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon