\(P=\left(1^2+2^2+...............+2015^2\right):\left(2^2+4^2+........+4030^2\right)\)

\(P=\left(1^2+2^2+............+2015^2\right):\left[\left(1.2\right)^2+\left(2.2\right)^2+.............+\left(2.2015\right)^2\right]\)

\(P=\left(1^2+2^2+........+2015^2\right):\left(1^2.2^2+2^2.2^2+...............+2015^2.2^2\right)\)

\(P=\left(1^2+2^2+......+2015^2\right):2^2.\left(1^2+2^2+.........+2015^2\right)\)

\(P=\left(1^2+2^2+........+2015^2\right).\frac{1}{2^2.\left(1^2+2^2+..............+2015^2\right)}\)

\(P=\frac{1^2+2^2+...............+2015^2}{2^2.\left(1^2+2^2+............+2015^2\right)}=\frac{1}{2^2}=\frac{1}{4}\)

Chúc bạn học tốt

Bài 1:

a) \(\frac{13.23-13}{26.21+26}\)=\(\frac{13.22}{26.22}\)=\(\frac{13}{26}\)=\(\frac{1}{2}\)

b) \(\frac{16.21-16}{4.19+4}\)=\(\frac{16.20}{4.20}\)=\(\frac{16}{4}\)= 4

\(\frac{5^3.90.4^3}{25^2.3^2.2^{13}}=\frac{5^3.2.3^2.5.2^6}{5^4.3^2.2^{13}}=\frac{5^4.2^7.3^2}{5^4.3^2.2^{13}}=\frac{2^7}{2^{13}}=\frac{128}{8192}=\frac{1}{64}\)

Bài 1a:

A = 2 + 2^2 + 2^3+ ...+ 2^100

2A = 2^2 + 2^3 + ...+ 2^101

2A - A = 2^2 + 2^3 + ...+ 2^101 - 2 - 2^2 - 2^3 - ... - 2^100

A = (2^2 - 2^2) + (2^3 - 2^3) + ... + (2^100 - 2^100) + (2^101 - 2)

A = 0+ 0+ 0 + ...+ 0 + 2^101 - 2

A = 2^101 - 2

Bài 2a:

A = 7^6 + 7^5 - 7^4

A = 7^4.(7^2 + 7 - 1)

A =7^4.(49 + 7 - 1)

A =7^4.(56 - 1)

A =7^4.55

A = 7^3.(7.11).5

A = 7^3.77.5 ⋮ 77 (đpcm)

1^2 +2^2+ ... +n^2 = n(n+1)(2n+1)/6