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Inverse of a Matrix

Sigma r =1 to...

Question

Sigma(r=1 to n) of [r/{(r^4)+(r^2)+1}] is equal to

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Solution

Solution is not incorrect. We can split the Summations as given above.

The Summation can be split over Numerator and Denominator and also to over each individual entity.

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Similar questions

	Column - I		Column - II
(A)	$\sum_{r = 1}^{n} r .^{n} C_{r}$ is equal to	(P)	$(n + 2) {.2}^{n - 1}$
(B)	$\sum_{r = 1}^{n + 1} r .^{n} C_{r - 1}$ is equal to	(Q)	$(n + 1) .^{2 n} C_{n}$
(C)	$\sum_{r = 0}^{n} (2 r + 1) .^{n} C_{r}$ is equal to	(R)	$(n + 1) {.2}^{n}$
(D)	$\sum_{r = 0}^{n} (2 r + 1) . (^{n} C_{r})^{2}$ is equal to	(S)	$n {.2}^{n - 1}$

r = n \sum r = 1 \frac{r^{4} + r^{2} + 1}{r^{4} + r} = \frac{675}{26},

n

2 \leq r \leq n, (\frac{n}{r}) + 2 (\frac{n}{r - 1}) + (\frac{n}{r - 2})

lim n \to \infty n \sum r = 1 {tan}^{- 1} (\frac{2 r}{1 - r^{2} + r^{4}})

\sum_{r = 1}^{n} r^{2} - \sum_{m = 1}^{n} \sum_{r = 1}^{m} r

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