If ∑n=2009∞1( n2009)=pq then which among the following option(s) is/are correct?(Note that (nk) is defined as n!k!(n−k)! and p and q are coprime integers)

Observe thatk+1k(1(n−1 k)−1(nk))=k+1k(nk)−(n−1 k)(nk)(n−1 k)=k+1k(n−1k−1)(nk)(n−1 k)=1( nk+1)Now apply this with k=2008 and sum across all n from 2009 to ∞. We get∑n=2009∞1( n2009)=20092008∑n=2009∞1(n−12008)−1( n2008)All terms from the sum on the right-hand-side cancel, except for the initial 1(20082008), which is equal to 1, So, we get ∑n=2009∞1( n2009)=20092008

If ∑n=2009∞1( n2009)=pq then which among the following option(s) is/are correct?(Note that (nk) is defined as n!k!(n−k)! and p and q are coprime integers)

Observe thatk+1k(1(n−1 k)−1(nk))=k+1k(nk)−(n−1 k)(nk)(n−1 k)=k+1k(n−1k−1)(nk)(n−1 k)=1( nk+1)Now apply this with k=2008 and sum across all n from 2009 to ∞. We get∑n=2009∞1( n2009)=20092008∑n=2009∞1(n−12008)−1( n2008)All terms from the sum on the right-hand-side cancel, except for the initial 1(20082008), which is equal to 1, So, we get ∑n=2009∞1( n2009)=20092008

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $\sum_{n = 2009}^{\infty} \frac{1}{(\begin{array}{l} n \\ 2009 \end{array})} = \frac{p}{q}$ then which among the following option(s) is/are correct?
(Note that $(\begin{array}{l} n \\ k \end{array})$ is defined as $\frac{n!}{k! (n - k)!}$ and p and q are coprime integers)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

p is divisible by 3

$p + q < 4006$

$p + q > 4006$

$p - q i s n e g a t i v e$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Observe that
$\frac{k + 1}{k} (\frac{1}{(\begin{array}{l} n - 1 \\ k \end{array})} - \frac{1}{(\begin{array}{l} n \\ k \end{array})}) = \frac{k + 1}{k} \frac{(\begin{array}{l} n \\ k \end{array}) - (\begin{array}{l} n - 1 \\ k \end{array})}{(\begin{array}{l} n \\ k \end{array}) (\begin{array}{l} n - 1 \\ k \end{array})} = \frac{k + 1}{k} \frac{(\begin{array}{l} n - 1 \\ k - 1 \end{array})}{(\begin{array}{l} n \\ k \end{array}) (\begin{array}{l} n - 1 \\ k \end{array})} = \frac{1}{(\begin{array}{l} n \\ k + 1 \end{array})}$
Now apply this with $k = 2008$ and sum across all $n$ from $2009$ to $\infty$ . We get
$\sum_{n = 2009}^{\infty} \frac{1}{(\begin{array}{l} n \\ 2009 \end{array})} = \frac{2009}{2008} \sum_{n = 2009}^{\infty} \frac{1}{(\begin{array}{l} n - 1 \\ 2008 \end{array})} - \frac{1}{(\begin{array}{l} n \\ 2008 \end{array})}$

All terms from the sum on the right-hand-side cancel, except for the initial $\frac{1}{(\begin{array}{l} 2008 \\ 2008 \end{array})}$ , which is equal to 1,
So, we get $\sum_{n = 2009}^{\infty} \frac{1}{(\begin{array}{l} n \\ 2009 \end{array})} = \frac{2009}{2008}$