【答案】
分析:(1)利用等差數列的中項公式列出關于a的等式,求出首項a,利用等差數列的前n項和公式列出關于k的等式,求出k的值.
(2)利用等差數列的前n項和公式求出S
n=n(n+1)得到
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,將其裂成兩項的差,利用裂項求和的方法求出和.
解答:解:(1)∵等差數列前三項為a,4,3a,
∴2×4=a+3a,
∴a=2,
公差d=4-2=2
又∵S
k=2550,
∴2k+
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×2=2550,
∴k
2+k-2550=0,
∴k=50或k=-51(不合,舍去),即k=50
(2)等差數列2,4,6,…的前n項和S
n=
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,即S
n=n(n+1)
于是
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=
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=
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-
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,
從而
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+
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+
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+…+
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=(1-
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)+(
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-
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)+(
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-
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)+…+(
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-
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)=1-
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=
點評:求數列的前n項和,應該先求出數列的通項,根據通項的特點選擇合適的求和方法.常用的求和方法有:公式法、錯位相減法、裂項相消法、倒序相加法、分組法.