Thu gọn tổng sau : A = 1 + 3 + 3^2 + 3^3 +…….+ 3^50 16/07/2021 Bởi Eden Thu gọn tổng sau : A = 1 + 3 + 3^2 + 3^3 +…….+ 3^50
${A}$ = ${1}$+${3}$+$3^{2}$+…+$3^{50}$ ${3A}$ = ${3}$ . ( ${1}$+${3}$+$3^{2}$+…+$3^{50}$ ) ${3A}$ = ${3}$+$3^{2}$+$3^{3}$…+$3^{51}$ ${3A – A}$ = ( ${3}$+$3^{2}$+$3^{3}$…+$3^{51}$ ) – ( ${1}$+${3}$+$3^{2}$+…+$3^{50}$ ) ${2A}$ = ${3}$+$3^{2}$+$3^{3}$…+$3^{51}$ – ${1}$-${3}$-$3^{2}$-…-$3^{50}$ ${2A}$ = $3^{51}$ – ${1}$ ${A}$ = $\frac{3^{51} – {1}}{2}$ $\text{Vậy A = $\frac{3^{51} – {1}}{2}$}$ Bình luận
A = 1 . 3 . `3^2` . `3^3` . ……. .`3^50`
A = `3^(0+1+2+….+50)`
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${A}$ = ${1}$+${3}$+$3^{2}$+…+$3^{50}$
${3A}$ = ${3}$ . ( ${1}$+${3}$+$3^{2}$+…+$3^{50}$ )
${3A}$ = ${3}$+$3^{2}$+$3^{3}$…+$3^{51}$
${3A – A}$ = ( ${3}$+$3^{2}$+$3^{3}$…+$3^{51}$ ) – ( ${1}$+${3}$+$3^{2}$+…+$3^{50}$ )
${2A}$ = ${3}$+$3^{2}$+$3^{3}$…+$3^{51}$ – ${1}$-${3}$-$3^{2}$-…-$3^{50}$
${2A}$ = $3^{51}$ – ${1}$
${A}$ = $\frac{3^{51} – {1}}{2}$
$\text{Vậy A = $\frac{3^{51} – {1}}{2}$}$