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MathematicsBinomial TheoremLogarithmic and Exponential seriesHard2 minPYQ_2019
MathematicsHardnumerical

Supposedetk=0nkk=0nCk nk2k=0nCk nkk=0nCk n3k=0,holds for some positive integern.Thenk=0nCk nk+1equals

Answer:
6.20
Solution:

As given k=0 n k k=0 n C   n k k 2 k=0 n C   n k k k=0 n C   n k 3 k =0
a k=0nk=nn+12
b k=0nCk nk2=k=0nk2-k+knCk
=k=0nk2-kCk n+k=0nkCk n
=k=0nkk-1nk.n-1k-1Ck-2 n-2+k=0nknkCk-1 n-1
=nn-1k=0nCk-2 n-2+nk=0nCk-1 n-1
=nn-12n-2+n2n-1=n2n-2n-1+2=nn+12n-2
rCr n=nCr-1 n-1 r=0nCr n=2n
c k=0nCk nk=k=0nkCk n=k=0nknkCk-1 n-1
=nk=0nCk-1 n-1=n2n-1
d k=0nCk n3k=C0 n+C13+C2 n32++Cn n3n n
=1+3n
=4n
now n n+1 2 n n+1 2 n2 n 2 n1 4 n =0

nn+122n-1-n2n+122n-3=0

22n-1-n22n-3=0
n=4
Now k=04Ck 4k+1=15k=04Ck+1 5
=1525-1Cr nr+1=Cr+1 n+1n+1
=315=6.20 r=1nCr=2r-1 n

Stream:JEE_ADVSubject:MathematicsTopic:Binomial TheoremSubtopic:Logarithmic and Exponential series
2mℹ️ Source: PYQ_2019

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