Prove by mathematical induction that 1x4 + 2x7 + 3x10 +... + n (3N + 1) = n (n + 1) &# 178;,

Prove by mathematical induction that 1x4 + 2x7 + 3x10 +... + n (3N + 1) = n (n + 1) &# 178;,

Analysis: 1 ° when n = 1, 1x4 = 1x2x2 obviously holds; 2 ° suppose that when n = k, the equation holds, that is, 1x4 + 2x7 + 3x10 +... + n (3K + 1) = K (K + 1) & # 178; then when n = K + 1, left = 1x4 + 2x7 + 3x10 +... + K (3K + 1) + (K + 1) [3 (K + 1) + 1] = K (K + 1) & # 178; + (K + 1) [3 (K + 1) + 1] = (K + 1) (K & # 178; + K + 3

We know that | m-2 | + (3-3n) ^ 2 = 0, and simplify 2x ^ (m-n + 1) y ^ 3-6y ^ (M + n) x ^ 2

Because | m-2 | + (3-3n) ^ 2 = 0, the two terms of left summation are nonnegative, and the sum is 0, which means that both are equal to 0
So m = 2, n = 1
2x^(m-n+1)y^3-6y^(m+n)x^2
=2x^(2-1+1)y^3-6y^(2+1)x^2
=2x^2y^3-6y^3x^2
=-4x^2y^3

Given | m-2 | + (3-3n) & sup2; = 0, simplify 2x ^ M-N + 1 y ^ 3-6y ^ m + n x & sup2;

|m-2|+(3-3n)²=0
If one is greater than 0, the other is less than 0
So both are equal to zero
So m-2 = 0, 3-3n = 0
m=2,n=1
Then M-N + 1 = 2
m+n=3
So the original formula = 2x & sup2; Y & sup3; - 6x & sup2; Y & sup3; = - 4x & sup2; Y & sup3;

1+4+7+10+13+.+{3n-2}=

1+4+7+10+13+.+{3n-2}
=[1+(3n-2)]×n÷2
=(3N & # 178; - n) / 2 {the square of 3N / 2 minus n}
This is the sum of arithmetic sequence, I hope to help you

The equation (1) x + y = 5 X-Y = 3 (2) M-N = 1 2m + 3N = 7 is solved by the method of addition, subtraction and elimination

(1) x+y=5 ①
x-y=3 ②
①+②＝2y＝8
y＝4
x＝1
（2）
m-n=1 ①
2m+3n=7 ②
①*3+②＝5m＝10
m＝2
n＝1

Let m.n be the two roots of the quadratic equation x ^ 2 + 3x-8 = 0, then m ^ 2 + 6m + 3N =?

x=m
Substituting
m²+3m-8=0
So M & # 178; + 6m = 3M + 8
By Weida theorem
m+n=-3
So the original formula = 3M + 8 + 3N
=3(m+n)+8
=3*(-3)+8
=-1

X & # 178; - 2mx-8m & # 178; = 0 and 20m & # 178; X & # 178; + 11mnx-3n & # 178; = 0 (solve the equation about x)

x²-2mx-8m²=0
(x+2m)(x-4m)=0,
X = - 2m or x = 4m
20m²x²+11mnx-3n²=0
(5mx-n)(4mx+3n)=0
X = n / 5m or x = - 3N / 4m

Find the solutions of these two quadratic equations of one variable: 6 (n + 1) & # 178; > n (n + 1) (2n + 1) 6N & # 178; > n (n-1) (2n-1)

A:
（1）
6(n+1)² ＞ n(n+1)(2n+1)
When n + 1 > 0, i.e. n > - 1:
6n+6>2n^2+n
2n^2-5n-6

√ m ^ 3N ^ 4 simplification

=mn²√m

Can 4m ^ 3N + 5MN ^ 3 be simplified

The original formula = Mn (4m & # 178; + 5N & # 185;)
If you don't understand this question, you can ask,