Tính các tổng sau: a,A=2+2 mũ2+2 mũ3+2 mũ4+…+2 mũ 100 b,C=1+5+5 mũ2+5 mũ3+…+1998 06/07/2021 Bởi Clara Tính các tổng sau: a,A=2+2 mũ2+2 mũ3+2 mũ4+…+2 mũ 100 b,C=1+5+5 mũ2+5 mũ3+…+1998
$a,A=2+2^{2}+2^{3}+2^{4}+…+2^{100}$ $2A=2^{2}+2^{3}+2^{4}+…+2^{100}+2^{101}$ $2A-A=(2^{2}+2^{3}+2^{4}+…+2^{100}+2^{101})-(2+2^{2}+2^{3}+2^{4}+…+2^{100})_{}$ $A=2^{101}-2_{}$ $b,C=`1+5+5^{2}+5^{3}+…+5^{1998}$ $5C=5+5^{2}+5^{3}+…+5^{1998}+5^{1999}$ $5C-C=(5+5^{2}+5^{3}+…+5^{1998}+5^{1999})-(1+5+5^{2}+5^{3}+…+5^{1998})_{}$ $4C=5^{1999}-1_{}$ $C=\frac{5^{1999}-1}{4}_{}$ Bình luận
a) A= $2^{1}$ + $2^{2}$ + ….+ $2^{100}$ =>2A = $2^{2}$ + $2^{3}$ + ….+ $2^{101}$ 2A – A =($2^{2}$ + $2^{3}$ + ….+ $2^{101}$)-($2^{1}$ + $2^{2}$ +….+ $2^{100}$ \ => A =$2^{101}$ -2 b) C= 1+5+$5^{2}$ +….+$5^{1998}$ 5C =5+$5^{2}$ +….+ $5^{1999}$ 5C-C= (5+$5^{2}$ +….+ $5^{1999}$ )-( 1+5 +….+ $5^{1998}$ ) 4C = $5^{1999}$ -1 => C=$\frac{5^{1999} -1}{4}$ Bình luận
$a,A=2+2^{2}+2^{3}+2^{4}+…+2^{100}$
$2A=2^{2}+2^{3}+2^{4}+…+2^{100}+2^{101}$
$2A-A=(2^{2}+2^{3}+2^{4}+…+2^{100}+2^{101})-(2+2^{2}+2^{3}+2^{4}+…+2^{100})_{}$
$A=2^{101}-2_{}$
$b,C=`1+5+5^{2}+5^{3}+…+5^{1998}$
$5C=5+5^{2}+5^{3}+…+5^{1998}+5^{1999}$
$5C-C=(5+5^{2}+5^{3}+…+5^{1998}+5^{1999})-(1+5+5^{2}+5^{3}+…+5^{1998})_{}$
$4C=5^{1999}-1_{}$
$C=\frac{5^{1999}-1}{4}_{}$
a) A= $2^{1}$ + $2^{2}$ + ….+ $2^{100}$
=>2A = $2^{2}$ + $2^{3}$ + ….+ $2^{101}$
2A – A =($2^{2}$ + $2^{3}$ + ….+ $2^{101}$)-($2^{1}$ + $2^{2}$ +….+ $2^{100}$ \
=> A =$2^{101}$ -2
b) C= 1+5+$5^{2}$ +….+$5^{1998}$
5C =5+$5^{2}$ +….+ $5^{1999}$
5C-C= (5+$5^{2}$ +….+ $5^{1999}$ )-( 1+5 +….+ $5^{1998}$ )
4C = $5^{1999}$ -1
=> C=$\frac{5^{1999} -1}{4}$