1×2+2×3+3×4+… +99×100．

1×2+2×3+3×4+… +99×100．

1×2+2×3+3×4+… +99×100，=1×（1+1）+2×（2+1）+3×（3+1）+… +98×（98+1）+99×（99+1），=12+1+22+2+32+3+… +982+98+992+99，=（12+22+32+… +982+992）+（1+2+3+… +98+99），=99×（99+1）×（2×99+1）÷6+...

The n power of 8 plus 1 power is equal to 16n-2 power. What is the n power equal to?

8^(n+1)=16^(n-2)
(2^3)^(n+1)=(2^4)^(n-2)
2^(3n+3)=2^(4n-8)
3n+3=4n-8
n=11

If 2 × 8N × 16N = 222, then n=______ ．

∵ 2 × 8N × 16N = 2 × 23n × 24N = 21 + 7n = 222; ∵ 1 + 7n = 22, the solution is n = 3

The fourth power of M - the fourth power of 16N

m4-16n4
=(m²)²-(4n²)²
=(m²+4n²)(m²-4n²)
=(m²+4n²)(m+2n)(m-2n)

The fourth power of M minus the fourth power of 16N is equal to?
Decomposition of factor

m^4-16n^4
=(m²+4n²)(m²-4n²)
=(m²+4n²)(m-2n)(m+2n)

The second power of M - the second power of 16N

The second power of M - the second power of 16N
=（m+4n）（m-4n）
If you don't understand this question, you can ask,

(5th power of m-5th power of 32n) / (4th power of 16n-4th power of M)=

(m-2n) (m ^ 4 + 4 + 2m (n + 4m; n \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(n + 4 Mn & # 178; + 8 N & # 179;)

Let a square matrix a satisfy a2-a-2e = 0, prove that a and a + 2E can be inverse, and find the inverse matrix of a and a + 2E

It is proved that: the ∵ square matrix a satisfies a2-a-2e = 0, ∵ a2-a = 2E, ∵ a × a − E2 = e, so a is invertible, and the inverse matrix is a − E2; the ∵ square matrix a satisfies a2-a-2e = 0, ∵ A2 = a + 2E, so a + 2e is invertible, and the inverse matrix is [a − E2] 2 = (a − E) 24

Let n-order square matrix a satisfy a ^ 2 = 3A, prove that a-4i is invertible, and obtain its inverse matrix

It is known that a ^ 2-3a = 0
So a (A-4E) + (A-4E) + 4E = 0
So (a + e) (A-4E) = - 4E
So A-4E is reversible and (A-4E) ^ - 1 = - 1 / 4 (a + e)

Linear algebra problem proves that if matrix A is invertible, then a can be expressed as the product of a series of elementary matrices
Thank you very much

Certificate:
If a is reversible, then the rank of a is n
So it can be transformed into standard form by elementary transformation, and p1p2... Psaq1q2... QT = E
PI (I = 1... S) is the elementary matrix for a row transformation, QJ (J = 1... T) is the elementary matrix for a column transformation
Because the inverse PI (I = 1... S) of PI and the inverse QJ (J = 1... T) of QJ are still elementary matrices
So a = PS... P2p1eq1q2... QT = PS... P2p1q1q2... QT
So a can be expressed as the product of a series of elementary matrices