Страница 232 номер 816, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением

1) \(S_{n} = \frac{a_{1} + a_{n}}{2} \cdot n\),
\(S_{m} = \dfrac{a_{1} + a_{m}}{2} \cdot m,\)
\(S_{n} = S_{m},\)
\(na_{1} + n\left(a_{1} + (n - 1)d\right) = ma_{1} + m\left(a_{1} + (m - 1)d\right),\)
\(a_{1}(n - m) = -a_{1}(n - m) + d(n - m) - d\left(n^{2} - m^{2}\right),\)

Так как \(n \neq m\), разделим на \(n - m\):

\(2a_{1} = d - d(n + m),\)
\(a_{1} + \dfrac{d}{2}(n + m - 1) = 0;\)
\(S_{n + m} = \dfrac{a_{1} + a_{n + m}}{2} \cdot (n + m) =\)
\(= \dfrac{a_{1} + a_{1} + (n + m - 1)d}{2} \cdot (n + m) =\)
\(= \left(a_{1} + \dfrac{d}{2}(n + m - 1)\right)(n + m) = 0;\)

2) \(S_{m} = \frac{a_{1} + a_{m}}{2} \cdot m = \frac{a_{1} + a_{1} + (m - 1)d}{2} \cdot m = \frac{2a_{1} + (m - 1)d}{2} \cdot m\),
\(S_{n} = \dfrac{a_{1} + a_{n}}{2} \cdot n = \dfrac{a_{1} + a_{1} + (n - 1)d}{2} \cdot n = \dfrac{2a_{1} + (n - 1)d}{2} \cdot n,\)
\(\dfrac{S_{m}}{S_{n}} = \dfrac{m^{2}}{n^{2}},\)
\(\dfrac{2a_{1} + (m - 1)d}{2a_{1} + (n - 1)d} \cdot \dfrac{m}{n} = \dfrac{m^{2}}{n^{2}},\)
\(\dfrac{2a_{1} + (m - 1)d}{2a_{1} + (n - 1)d} = \dfrac{m}{n},\)
\(n\left(2a_{1} + (m - 1)d\right) = m\left(2a_{1} + (n - 1)d\right),\)
\(2a_{1}n + nmd - nd = 2a_{1}m + nmd - md,\)
\(2a_{1}(n - m) = d(n - m).\)

При \(n = m\) равенство \(\frac{a_{m}}{a_{n}} = \frac{2m - 1}{2n - 1}\) выполняется.

При \(n \neq m\):

\(2a_{1} = d,\)
\(\dfrac{a_{m}}{a_{n}} = \dfrac{a_{1} + d(m - 1)}{a_{1} + d(n - 1)} = \dfrac{a_{1} + 2a_{1}(m - 1)}{a_{1} + 2a_{1}(n - 1)} =\)
\(= \dfrac{a_{1}(1 + 2m - 2)}{a_{1}(1 + 2n - 2)} = \dfrac{2m - 1}{2n - 1};\)

3) \(S_{n + 3} = 3S_{n + 2} - 3S_{n + 1} + S_{n}\)
\(S_{n} = \dfrac{a_{1} + a_{n}}{2}n, a_{n} = a_{1} + (n - 1)d,\)
\(S_{n + 1} = \dfrac{a_{1} + a_{n + 1}}{2}(n + 1), a_{n + 1} = a_{1} + nd,\)
\(S_{n + 2} = \dfrac{a_{1} + a_{n + 2}}{2}(n + 2), a_{n + 2} = a_{1} + (n + 1)d,\)
\(S_{n + 3} = \dfrac{a_{1} + a_{n + 3}}{2}(n + 3), a_{n + 3} = a_{1} + (n + 2)d,\)
\(3S_{n + 2} - 3S_{n + 1} + S_{n} =\)
\(= 3 \cdot \dfrac{a_{1} + a_{n + 2}}{2}(n + 2) - 3 \cdot \dfrac{a_{1} + a_{n + 1}}{2}(n + 1) +\)
\(+\dfrac{a_{1} + a_{n}}{2}n = \dfrac{3}{2}\left(2a_{1} + (n + 1)d\right)(n + 2) - \dfrac{3}{2}\left(2a_{1} + nd\right)(n + 1) +\)
\(+\dfrac{1}{2}\left(2a_{1} + (n - 1)d\right)n =\)
\(= \dfrac{\begin{gathered}3\left(2na_{1} + 4a_{1} + (n + 1)(n + 2)d\right) -\\ - 3\left(2na_{1} + 2a_{1} + n(n + 1)d\right) + 2na_{1} + n(n - 1)d\end{gathered}}{2} =\)
\(= \dfrac{\begin{gathered}12a_{1} + 3(n + 1)(n + 2)d - 6a_{1} -\\ - 3n(n + 1)d + 2na_{1} + n(n - 1)d\end{gathered}}{2} =\)
\(= \dfrac{a_{1}(6 + 2n) + d\left(3(n + 1)(n + 2) - 3n(n + 1) + n(n - 1)\right)}{2} =\)
\(= \dfrac{2a_{1}(3 + n) + d\left(n^{2} + 5n + 6\right)}{2} =\)
\(= \dfrac{2a_{1}(3 + n) + (3 + n)(n + 2)d}{2} =\)
\(= \dfrac{(3 + n)\left(2a_{1} + (n + 2)d\right)}{2} = S_{n + 3};\)

4) \(S_{3n} = 3\left(S_{2n} - S_{n}\right)\)
\(3\left(S_{2n} - S_{n}\right) = 3\left(\dfrac{a_{1} + a_{2n}}{2} \cdot 2n - \dfrac{a_{1} + a_{n}}{2} \cdot n\right) =\)
\(= 3\left(\left(a_{1} + a_{1} + (2n - 1)d\right)n - \dfrac{a_{1} + a_{1} + (n - 1)d}{2} \cdot n\right) =\)
\(= 3\left(2na_{1} + \left(2n^{2} - n\right)d - \dfrac{2na_{1} + \left(n^{2} - n\right)d}{2}\right) =\)
\(= 3\left(\dfrac{4na_{1} + \left(4n^{2} - 2n\right)d - 2na_{1} - \left(n^{2} - n\right)d}{2}\right) =\)
\(= \dfrac{3}{2}n\left(2a_{1} + (3n - 1)d\right);\)
\(S_{3n} = \dfrac{a_{1} + a_{3n}}{2} \cdot 3n = \dfrac{a_{1} + a_{1} + (3n - 1)d}{2} \cdot 3n =\)
\(= \dfrac{3}{2}n\left(2a_{1} + (3n - 1)d\right);\)

Левая и правая части равны.