If an=r=0∑n nCr1 then r=0∑n nCrr equals

We have an=r=0∑nnCr1 =r=0∑2+1n nCr1+nCn−r1 ⇒an=r=0∑2n+1 nCr2=2r=0∑2n+1 nCr1 ∴r=0∑nnCr1 =r=0∑2n+1 nCrr+(n−r) ⇒nr=0∑nnCr1 =n(2an )=2nan

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If { a }_{ n }= _{ r=0 }^{ n }{ { 1 }{ { _{ }^{ n }{ C } }_{

Question

If $a_{n} = r = 0 \sum n \frac{1}{_{n} C _{r}}$ then $r = 0 \sum n \frac{r}{_{n} C _{r}}$ equals

(n - 1) a_{n}

n a_{n}

\frac{1}{2} n a_{n}

\frac{( n - 1 )}{2} a_{n}

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Oliver Discussed

a_{n} = r = 0 \sum n \frac{1}{_{n} C _{r}}

then

r = 0 \sum n \frac{r}{_{n} C _{r}}

equals

12 mins ago

Discuss this question LIVE

12 mins ago

We have

a_{n} = r = 0 \sum n \frac{1}{^{n} C _{r}} = r = 0 \sum \frac{n}{2 + 1} \frac{1}{^{n} C _{r}} + \frac{1}{^{n} C _{n - r}}

\Rightarrow a_{n} = r = 0 \sum \frac{n + 1}{2} \frac{2}{^{n} C _{r}} = 2 r = 0 \sum \frac{n + 1}{2} \frac{1}{^{n} C _{r}}

∴ r = 0 \sum n \frac{1}{^{n} C _{r}} = r = 0 \sum \frac{n + 1}{2} \frac{r + ( n - r )}{^{n} C _{r}}

\Rightarrow n r = 0 \sum n \frac{1}{^{n} C _{r}} = n (\frac{a _{n}}{2}) = \frac{n}{2} a_{n}

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Question Text	If $a_{n} = r = 0 \sum n \frac{1}{_{n} C _{r}}$ then $r = 0 \sum n \frac{r}{_{n} C _{r}}$ equals
Answer Type	Text solution:1
Upvotes	150