If ∑ r = 0 n ( r n ) r = ∑ r = 0 n 2 ( r n ) n 2 − 4 n + 6 , then which values of n are possible?

By replacing r with n − r , symmetry gives ∑ r / ( r n ) = 2 n ∑ 1/ ( r n ) . Therefore n = n 2 − 4 n + 6 , or ( n − 2 ) ( n − 3 ) = 0 . Hence n = 2 , 3 .

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Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremProperties of Binomial coefficientsMedium2 minQB

MathematicsMediummultiple choice

If $\sum_{r = 0}^{n} \frac{r}{( r n )} = \sum_{r = 0}^{n} \frac{n ^{2} - 4 n + 6}{2 ( r n )}$ , then which values of $n$ are possible?

Options:(select one or more)

Answer:

B, C

Solution:

By replacing $r$ with $n - r$ , symmetry gives $\sum r / (r n) = \frac{n}{2} \sum 1/ (r n)$ . Therefore $n = n^{2} - 4 n + 6$ , or $(n - 2) (n - 3) = 0$ . Hence $n = 2, 3$ .

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:Properties of Binomial coefficients

⏱ 2mℹ️ Source: QB

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