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    等式左右两边同时对i求和,用英语怎么说

        发布时间:2019-09-17

        21;
        alpha2=5,22:
        V=[413:
        C =
        5;
        syms C
        B=(-alpha1-alpha2+sqrt((alpha1+alpha2)^2-4*beta*(log(C)-log(r)-log(prod(V)))))/,36,42;
        alpha1=2;
        beta=4;
        C=double(solve(B))
        结果;
        r=10;beta,78],225可这样

        回复:

        In recent years, more emphasis on capacity. The use of recursive formula for general term for the analysis to be passed out for a few and, which requires us to address a number of different issues out to master a certain sum of methods and techniques. But the students to deal with such questions is a commonly used method of mathematical induction and the general inequality of the zoom is often done on the half-way up, purposeful, it is the students have mastered the method. Inspired by the author in order to form series and prove that the inequality is not only flexibility in the use of sum test methods, fast solution of the problem was, but also to examine the proof of the zoom
        Skills, to enable students to permit such practice in the solution when communication and flexibility in the use of the school series knowledge; to be passed to a reasonable zoom;, can be transformed into the form of summation series to prove that inequality is the author of this article attempts to explore problems. Analysis of this paper gives several examples of effective solutions, the college entrance examination in volume, "Ben to "These" target ", the sum of the test series of new.就是这样了Series summation problem is that the basic contents of series one of the hot spots and focus on college entrance examination, clever series will be passed zoom appropriate, the sum of its easy

        回复:

        (1+1)^2+(2+1)^2+(3+1)^2+…+(n+1)2^=1^2+2^2+3^2+…+n^2+2(1+2+…+n)+n由题意,
        则(n+1)^2=1+2(1+2+3+…+n)+n,
        即2^2+3^2+4^2+…+(n+1)^2=1^2+2^2+3^2+…+n^2+2(1+2+…+n)+n,解答如下
        把已知的式子左右分别相加得,
        即2(1+2+3+4+…+n)=n2+n
        ∴1+2+3+4+5+6+…+n=n(n+1)/

        回复:

        有两个等式,左右两边分别相加后,仍然相等,消去法解应用题的做法。

        回复:

        Series summation problem is that the basic contents of series one of the hot spots and focus on college entrance examination. In recent years, the college entrance examination in volume, the sum of the test series of new, more ...

        回复:

        由题意,解答如下 把已知的式子左右分别相加得:(1+1)^2+(2+1)^2+(3+1)^2+…+(n+1)2^=1^2+2^2+3^2+…+n^2+2(1+2+…+n)+n, 即2^2+3^2+4^2+…+(n+1)^2=1^2+2^2+3^2+…+n^2+2(1+2+…+n)+n, 则(n+1)^2=1+2(1+2+3+…+n)+n, 即2(1+2+3+...

        回复:

        不等式的两边同时加上或减去同一个数,不等号方向不变; 不等式的两边同时乘以或者除以一个正数,不等号的方向不变; 不等式的两边同时乘以或者除以一个负数,不等号的方向改变。

        回复:

        两同向不等式间只能相加,不能相减;只能相乘,不能相除。

        回复:

        大约结论是错的。没有这个等式,可能只对特殊情况成立。 这个等式解释是(右边)一些分数和的平均值等于分子的和除以分母的和(左边) 显然不是真命题

        回复:

        可这样: V=[413,225,21,22,42,36,78]; alpha1=2; alpha2=5; beta=4; r=10; syms C B=(-alpha1-alpha2+sqrt((alpha1+alpha2)^2-4*beta*(log(C)-log(r)-log(prod(V)))))/beta; C=double(solve(B)) 结果: C = 5.0632e+13

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