Given the standard result \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\), find the value of \(\sum_{r=1}^{10} r^2\).
Cambridge International AS Level · Mathematics - Further (9231)
級數求和:练习题
5 道选择题即时批改,另有 4 道文字题附完整解题步骤,全部围绕「級數求和」。
By using partial fractions, the general term of a series is given by \(u_r = \frac{1}{(r+1)(r+2)} = \frac{1}{r+1} - \frac{1}{r+2}\). Find the sum of the first \(n\) terms, \(S_n = \sum_{r=1}^{n} u_r\), and determine the sum to infinity \(S_{\infty}\).
Use the standard results for \(\sum r, \sum r^2\) and \(\sum r^3\) to find the value of \(\sum_{r=n+1}^{2n} r(r+1)\) in terms of \(n\).
Using the standard formula for the sum of the first \(n\) positive integers, evaluate the sum \(\sum_{r=1}^{20} (2r + 1)\).
By using the method of differences on the identity \(\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{2r+1}{r^2(r+1)^2}\), find the sum to infinity of the series \(\sum_{r=1}^{\infty} \frac{2r+1}{r^2(r+1)^2}\).
Using the standard formula for the sum of the first \( n \) natural numbers, find the value of \( \sum_{r=1}^{20} (2r + 1) \).
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Find the sum of the series \(\sum_{r=1}^n \frac{1}{(2r-1)(2r+1)}\) by using the method of differences.
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Find the sum to infinity of the series \(\sum_{r=1}^{\infty} \frac{1}{r(r+1)(r+2)}\) by using the method of partial fractions and differences.
先自己写一遍答案,再对照解题步骤。
(a) Show that \(\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}\).
(b) Hence, use the method of differences to show that \(\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1}\).
(c) State whether the series \(\sum_{r=1}^{\infty} \frac{1}{r(r+1)}\) is convergent and find its sum to infinity.
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