Đề cương ôn tập cuối kì 1 Toán 6 Kết nối tri thức cấu trúc mới (Tự luận) có đáp án - Phần 3

a) Ta có \(A = 5 + {5^2} + {5^3} + ... + {5^{100}}\) nên \(5A = 5 \cdot \left( {5 + {5^2} + {5^3} + ... + {5^{100}}} \right) = {5^2} + {5^3} + {5^4} + ... + {5^{101}}\).

Suy ra \(5A - A = \left( {{5^2} + {5^3} + {5^4} + ... + {5^{101}}} \right) - \left( {5 + {5^2} + {5^3} + ... + {5^{100}}} \right)\)

Do đó \(4A = {5^{101}} - 5\) nên \(A = \frac{{{5^{101}} - 5}}{4}.\)

b) Ta có \(B = 1 + {3^2} + {3^4} + ... + {3^{100}}\) nên \({3^2} \cdot B = {3^2} \cdot \left( {1 + {3^2} + {3^4} + ... + {3^{100}}} \right) = {3^2} + {3^4} + {3^6} + ... + {3^{102}}\).

Suy ra \[{3^2}B - B = \left( {{3^2} + {3^4} + {3^6} + ... + {3^{102}}} \right) - \left( {1 + {3^2} + {3^4} + ... + {3^{100}}} \right)\]

Do đó \[8B = {3^{102}} - 1\] nên \[B = \frac{{{3^{102}} - 1}}{8}.\]

c) Ta có \[C = 2 - {2^2} + {2^3} - {2^4} + ... - {2^{100}}\] nên

\(2C = 2 \cdot \left( {2 - {2^2} + {2^3} - {2^4} + ... + {2^{99}} - {2^{100}}} \right) = {2^2} - {2^3} + {2^4} - {2^5} + ... + {2^{100}} - {2^{101}}\)

Suy ra \[2C + C = \left( {{2^2} - {2^3} + {2^4} - {2^5} + ... + {2^{100}} - {2^{101}}} \right) + \left( {2 - {2^2} + {2^3} - {2^4} + ... + {2^{99}} - {2^{100}}} \right)\]

Do đó \[3C = 2 - {2^{101}}\] nên \[C = \frac{{2 - {2^{101}}}}{3}.\]

d) Ta có \(D = 1 \cdot 2 + 2 \cdot 3 + ... + 99 \cdot 100\) nên \(3D = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 3 + ... + 99 \cdot 100 \cdot 3\)

Suy ra \(3D = 1 \cdot 2 \cdot \left( {3 - 0} \right) + 2 \cdot 3 \cdot \left( {4 - 1} \right) + ... + 99 \cdot 100 \cdot \left( {101 - 98} \right)\)

\[ = 1 \cdot 2 \cdot 3 - 1 \cdot 2 \cdot 0 + 2 \cdot 3 \cdot 4 - 2 \cdot 3 \cdot 1 + ... + 99 \cdot 100 \cdot 101 - 99 \cdot 100 \cdot 98\]

\[ = \left( {1 \cdot 2 \cdot 3 - 2 \cdot 3 \cdot 1} \right) + \left( {2 \cdot 3 \cdot 4 - 3 \cdot 4 \cdot 2} \right) + ... + \left( {98 \cdot 99 \cdot 100 - 99 \cdot 100 \cdot 98} \right) + \left( {99 \cdot 100 \cdot 101 - 1 \cdot 2 \cdot 0} \right)\]

\[ = 99 \cdot 100 \cdot 101 - 0 = 99 \cdot 100 \cdot 101\].

Do đó \(D = \frac{{99 \cdot 100 \cdot 101}}{3} = 33 \cdot 100 \cdot 101 = 333\,\,300\).

e) Ta có \(E = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + 98 \cdot 99 \cdot 100\) nên \(4E = 1 \cdot 2 \cdot 3 \cdot 4 + 2 \cdot 3 \cdot 4 \cdot 4 + ... + 98 \cdot 99 \cdot 100 \cdot 4\)

Suy ra \(4E = 1 \cdot 2 \cdot 3 \cdot 4 + 2 \cdot 3 \cdot 4 \cdot \left( {5 - 1} \right) + ... + 98 \cdot 99 \cdot 100 \cdot \left( {101 - 97} \right)\)

\( = 1 \cdot 2 \cdot 3 \cdot 4 + 2 \cdot 3 \cdot 4 \cdot 5 - 2 \cdot 3 \cdot 4 \cdot 1 + ... + 98 \cdot 99 \cdot 100 \cdot 101 - 98 \cdot 99 \cdot 100 \cdot 97\)

\( = \left( {1 \cdot 2 \cdot 3 \cdot 4 - 2 \cdot 3 \cdot 4 \cdot 1} \right) + \left( {2 \cdot 3 \cdot 4 \cdot 5 - 3 \cdot 4 \cdot 5 \cdot 2} \right) + ...\left( {97 \cdot 98 \cdot 99 \cdot 100 - 98 \cdot 99 \cdot 100 \cdot 97} \right) + 98 \cdot 99 \cdot 100 \cdot 101\)

\( = 98 \cdot 99 \cdot 100 \cdot 101\)

Do đó \(E = \frac{{98 \cdot 99 \cdot 100 \cdot 101}}{4} = 24\,\,497\,\,550.\)

f) \(F = {1^2} + {2^2} + {3^2} + ... + {99^2} + {100^2}\)

\( = 1 \cdot \left( {2 - 1} \right) + 2 \cdot \left( {3 - 1} \right) + 3 \cdot \left( {4 - 1} \right) + ... + 99 \cdot \left( {100 - 1} \right) + 100 \cdot \left( {101 - 1} \right)\)

\[ = 1 \cdot 2 - 1 \cdot 1 + 2 \cdot 3 - 2 \cdot 1 + 3 \cdot 4 - 3 \cdot 1 + ... + 99 \cdot 100 - 99 \cdot 1 + 100 \cdot 101 - 100 \cdot 1\]

\[ = 1 \cdot 2 - 1 + 2 \cdot 3 - 2 + 3 \cdot 4 - 3 + ... + 99 \cdot 100 - 99 + 100 \cdot 101 - 100\]

\[ = \left( {1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 - 3 + ... + 99 \cdot 100 + 100 \cdot 101} \right) - \left( {1 + 2 + ... + 99 + 100} \right)\]

\[ = \frac{{100 \cdot 101 \cdot 102}}{3} - \frac{{100 \cdot \left( {100 + 1} \right)}}{2}\]                   (tương tự câu d)

\[ = \frac{{100 \cdot 101 \cdot 102}}{3} - \frac{{100 \cdot 101}}{2}\]

\[ = \frac{{100 \cdot 101 \cdot 102 \cdot 2 - 100 \cdot 101 \cdot 3}}{6}\]

\( = \frac{{100 \cdot 101 \cdot 201}}{6} = 338\,\,350.\)

g) \(G = 1 \cdot 99 + 2 \cdot 98 + 3 \cdot 97 + ... + 98 \cdot 2 + 99 \cdot 1\)

\( = 1 \cdot 99 + 2 \cdot \left( {99 - 1} \right) + 3 \cdot \left( {99 - 2} \right) + ... + 98 \cdot \left( {99 - 97} \right) + 99 \cdot \left( {99 - 98} \right)\)

\[ = 1 \cdot 99 + 2 \cdot 99 - 2 \cdot 1 + 3 \cdot 99 - 3 \cdot 2 + ... + 98 \cdot 99 - 98 \cdot 97 + 99 \cdot 99 - 99 \cdot 98\]

\[ = \left( {1 \cdot 99 + 2 \cdot 99 + 3 \cdot 99 + ... + 98 \cdot 99 + 99 \cdot 99} \right) - \left( {2 \cdot 1 + 3 \cdot 2 + ... + 98 \cdot 97 + 99 \cdot 98} \right)\]

\[ = 99 \cdot \left( {1 + 2 + 3 + ... + 98 + 99} \right) - \frac{{98 \cdot 99 \cdot 100}}{3}\]                   (tương tự câu d)

\[ = 99 \cdot \frac{{99 \cdot \left( {99 + 1} \right)}}{2} - 98 \cdot 33 \cdot 100\]

\[ = 99 \cdot 99 \cdot 50 - 98 \cdot 33 \cdot 100 = 166\,\,650.\]

h) \(H = 1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + ... + 97 \cdot 99 + 98 \cdot 100\)

 \( = 1 \cdot \left( {2 + 1} \right) + 2 \cdot \left( {3 + 1} \right) + 3 \cdot \left( {4 + 1} \right) + ... + 97 \cdot \left( {98 + 1} \right) + 98 \cdot \left( {99 + 1} \right)\)

 \( = 1 \cdot 2 + 1 \cdot 1 + 2 \cdot 3 + 2 \cdot 1 + 3 \cdot 4 + 3 \cdot 1 + ... + 97 \cdot 98 + 97 \cdot 1 + 98 \cdot 99 + 98 \cdot 1\)

 \[ = \left( {1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + 97 \cdot 98 + 98 \cdot 99} \right) + \left( {1 + 2 + 3 + ... + 97 + 98} \right)\]

 \[ = \frac{{98 \cdot 99 \cdot 100}}{3} + \frac{{98 \cdot \left( {98 + 1} \right)}}{2}\]           (tương tự câu d)

 \[ = 98 \cdot 33 \cdot 100 + 49 \cdot 99 = 328\,\,251.\]