高中数学数列大题怎么写求解答谢谢 10
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以<> 表示下标。
a<n> = a<1> + (n-1)d, a<n+1> = a<1> + nd
b<n> = a<n> + a<n+1> = 2a<1> + (2n-1)d
(1) b<1> = 2a<1> + d = 36
b<3> = 2a<1> + 5d = 28
联立解得 d = -2, a<1> = 19, 则 a<n> = 19 - 2(n-1) = 21-2n
(2) a<1> = 2, b<n> = 2n+1, 则 b<1> = a<1> +a<2> = 3,
a<2> = 1, d = a<2>-a<1> = 1-2 = -1,
a<n> = a<1> + (n-1)d = 2-(n-1) = 3-n
S<n> = (1/2)n[a<1>+a<n>] = (1/2)n(2+3-n) = (1/2)n(5-n)
则 S<2n+1> = a<1>+a<2>+ ... +a<n>
+ a<n+1> +a<n+2>+ ... +a<2n>+ a<2n+1>
= a<1>+a<2>+ ... +a<n> + (a<n>+a<1>) + (a<n>+a<2>)
+ ... + (a<n>+a<n>) + (2a<n>+a<1>)
= 2S<n> + (n+2)a<n> + a<1>
= n(5-n) + (n+2)(3-n) + 2 = 2(1-n)(4+n)
a<n> = a<1> + (n-1)d, a<n+1> = a<1> + nd
b<n> = a<n> + a<n+1> = 2a<1> + (2n-1)d
(1) b<1> = 2a<1> + d = 36
b<3> = 2a<1> + 5d = 28
联立解得 d = -2, a<1> = 19, 则 a<n> = 19 - 2(n-1) = 21-2n
(2) a<1> = 2, b<n> = 2n+1, 则 b<1> = a<1> +a<2> = 3,
a<2> = 1, d = a<2>-a<1> = 1-2 = -1,
a<n> = a<1> + (n-1)d = 2-(n-1) = 3-n
S<n> = (1/2)n[a<1>+a<n>] = (1/2)n(2+3-n) = (1/2)n(5-n)
则 S<2n+1> = a<1>+a<2>+ ... +a<n>
+ a<n+1> +a<n+2>+ ... +a<2n>+ a<2n+1>
= a<1>+a<2>+ ... +a<n> + (a<n>+a<1>) + (a<n>+a<2>)
+ ... + (a<n>+a<n>) + (2a<n>+a<1>)
= 2S<n> + (n+2)a<n> + a<1>
= n(5-n) + (n+2)(3-n) + 2 = 2(1-n)(4+n)
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