如何求等差数列之和

部分 1

部分 1 的 3:评估数列

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1
确定数列是等差数列。等差数列是一组有规律的数字,其中各数字的增量是一个常数。
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本文所述方法仅适用于等差数列。
要确定数列是否是等差数列,你可以计算前面几个数字之间的差值和最后几个数字之间的差值。等差数列的差值应始终相等。
例如,数列10, 15, 20, 25, 30是一个等差数列,因为各项之间的差值等于常数(5)。
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确定数列的项数。每个数字构成一项。如果数列只包含列出的几个数字,你可以数一数共有多少项。否则,在知道首项、末项,以及被称为公差的各项之差的情况下,你可以使用公式来算出项数。我们可以使用变量n{displaystyle n}来代表这个数字。
例如,如果你要计算数列10, 15, 20, 25, 30之和,则n=5{displaystyle n=5},因为数列共有5项。

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3
确定数列的首项和末项。要计算等差数列之和,你必须知道这两个数字。第一个数字常常为1,但也并不一定。我们可以设变量a1{displaystyle a_{1}}等于数列首项,变量an{displaystyle a_{n}}等于数列末项。
例如,在数列10, 15, 20, 25, 30中,a1=10{displaystyle a_{1}=10},而an=30{displaystyle a_{n}=30}。

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部分 2

部分 2 的 3:计算总和

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列出计算等差数列之和的公式。公式为Sn=n(a1+an2){displaystyle S_{n}=n({frac {a_{1}+a_{n}}{2}})},其中Sn{displaystyle S_{n}}等于数列之和。[2]
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注意,此公式表明等差数列之和等于首项和末项的平均数乘以项数。[3]
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将变量n{displaystyle n}、a1{displaystyle a_{1}}和an{displaystyle a_{n}}代入公式中。确保代入步骤正确。
例如,如果数列有5项,首项为10,末项为30,则代入后公式变成:Sn=5(10+302){displaystyle S_{n}=5({frac {10+30}{2}})}。

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计算首项和末项的平均数。将两个数字相加,然后除以2。
例如:Sn=5(402){displaystyle S_{n}=5({frac {40}{2}})}Sn=5(20){displaystyle S_{n}=5(20)}

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用平均数乘以数列的项数。这样就算出了等差数列之和。
例如:Sn=5(20){displaystyle S_{n}=5(20)}Sn=100{displaystyle S_{n}=100}因此,数列10, 15, 20, 25, 30之和等于100。

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部分 3

部分 3 的 3:完成例题

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1
计算1到500之间所有数字之和。考虑所有的连续整数。
确定数列的项数n{displaystyle n}。由于需要考虑500以内的所有连续整数,因此n=500{displaystyle n=500}。
确定数列的首项a1{displaystyle a_{1}}和末项an{displaystyle a_{n}}。由于数列是从1到500,所以a1=1{displaystyle a_{1}=1},而an=500{displaystyle a_{n}=500}。
计算a1{displaystyle a_{1}}和an{displaystyle a_{n}}的平均数:1+5002=250.5{displaystyle {frac {1+500}{2}}=250.5}。
用平均数乘以n{displaystyle n}:250.5×500=125,250{displaystyle 250.5times 500=125,250}。

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求下述等差数列之和。数列的首项为3。数列的末项为24。公差为7。
确定数列的项数n{displaystyle n}。由于数列的第一项为3,最后一项为24,而每一项比前一项大7,所以这个数列是3, 10, 17, 24。以上推论是根据公差的定义得出,公差即数列中各项与前一项之差。[4]
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这意味着n=4{displaystyle n=4}

确定数列的首项a1{displaystyle a_{1}}和末项an{displaystyle a_{n}}。由于数列是从3到24,所以a1=3{displaystyle a_{1}=3},而an=24{displaystyle a_{n}=24}。
计算a1{displaystyle a_{1}}和an{displaystyle a_{n}}的平均数:3+242=13.5{displaystyle {frac {3+24}{2}}=13.5}。
用平均数乘以n{displaystyle n}:13.5×4=54{displaystyle 13.5times 4=54}。

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3
解以下问题。陈静在一年的第一周存了5元钱。在这一年中剩下的时间里,她每周会比前一周多存5元钱。年末时,陈静共存了多少钱?
确定数列的项数n{displaystyle n}。由于陈静存了1年,而1年有52周,所以n=52{displaystyle n=52}。
确定数列的首项a1{displaystyle a_{1}}和末项an{displaystyle a_{n}}。她存的第一笔钱金额为5元,所以a1=5{displaystyle a_{1}=5}。她在这一年最后一周存的金额可以计算得出,5×52=260{displaystyle 5times 52=260}。因此,an=260{displaystyle a_{n}=260}。
计算a1{displaystyle a_{1}}和an{displaystyle a_{n}}的平均数:5+2602=132.5{displaystyle {frac {5+260}{2}}=132.5}。
用平均数乘以n{displaystyle n}:135.5×52=7,046{displaystyle 135.5times 52=7,046}。所以,她在年末时共存了7,046元。

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