Toán Lớp 6: thu gọn tổng sau a 1+2+2^2+2^3+-.+2^100 b 1+a+a^2+a^3+-.+a^n

Question

Toán Lớp 6:  thu gọn tổng sau a 1+2+2^2+2^3+.....+2^100 b 1+a+a^2+a^3+.....+a^n

thudan · Answer

a ) Gọi 1 + 2 + 2^2 + 2^3 + ... + 2^100 l&agrave; S   => 2S = 2 + 2^2 + 2^3 + 2^4 + ... + 2^101   => 2S - A = ( 2 + 2^2 + 2^3 + 2^4 + ... + 2^101 ) - ( 1 + 2 + 2^2 + 2^3 + ... + 2^100 )   => S = 2^101 - 1   Vậy 1 + 2 + 2^2 + 2^3 + ... + 2^100 = 2^101 - 1   b ) Gọi 1 + a + a^2 + a^3 + ... + a^n l&agrave; M   => 2M = a + a^2 + a^3 + a^4 + ... + 1^{n + 1}   => 2M - M = ( a + a^2 + a^3 + a^4 + ... + a^{n + 1} ) - ( 1 + a + a^2 + a^3 + ... + a^n )   => M = a^{n + 1} - 1   Vậy 1 + a + a^2 + a^3 + ... + a^n = a^{n + 1} - 1    #dtkc

thanhthuythu · Answer

a) 1 + 2 + $2^{2}$ + $2^{3}$ +...+ $2^{100}$    Đặt A = 1 + 2 + $2^{2}$ + $2^{3}$ +...+ $2^{100}$    2A = 2 + $2^{2}$ + $2^{3}$ + $2^{4}$ ...+ $2^{101}$    2A - A = (2 + $2^{2}$ + $2^{3}$ + $2^{4}$ ...+ $2^{101}$) - (1 + 2 + $2^{2}$ + $2^{3}$ +...+ $2^{100}$)   A = $2^{101}$ - 1   b) 1 + a + $a^{2}$ + $a^{3}$ +...+ $a^{n}$   Đặt B = 1 + a + $a^{2}$ + $a^{3}$ +...+ $a^{n}$   2B = a + $a^{2}$ + $a^{3}$ + $a^{4}$ + ...+ $a^{n+1}$   2B - B = (a + $a^{2}$ + $a^{3}$ + $a^{4}$ + ...+ $a^{n+1}$) - (1 + a + $a^{2}$ + $a^{3}$ +...+ $a^{n}$)   B = $a^{n+1}$ - 1   ---------------------------------------    Xin hay nhất ạ!!~