Tính giá trị biểu thức :
E = $2$ + $2^{2}$ + $2^{3}$ + $2^{4}$ + … + $2^{100}$
F = $1$ + $3^{1}$ + $3^{2}$ + $3^{3}$ + … + $3^{100}$
G = $5$ + $5^{3}$ + $5^{5}$ + $5^{7}$ + … + $5^{99}$
H = $(1 + 2 + 3 + … + 100)$. ($1^{2}$ + $2^{2}$ + $3^{2}$ + … + $100^{2}$).$(65.111 – 13.15.37)$
+) 2E = $2^{2}$ +$2^{3}$ + …. +$2^{101}$
=> 2E – E =$2^{2}$ +$2^{3}$ + …. +$2^{101}$ – 2 – $2^{2}$ – $2^{3}$ – …. -$2^{101}$
=> E = $2^{101}$ – 2
+) 3F = 3 + $3^{2}$ +$3^{3}$ + …. +$3^{101}$ –
=> 3F – F = 3 + $3^{2}$ +$3^{3}$ + …. +$3^{101}$ – 1 – $3^{1}$ +$3^{2}$ + …. +$3^{100}$
=> 2F = $3^{101}$ + 2
=> F = $\frac{3^{101}}{2}$ + 1
+) B.$5^{2}$ = $5^{3}$ + $5^{5}$ + … + $5^{101}$
=> 25B – B = $5^{3}$ + $5^{5}$ + … + $5^{101}$ – 5 – $5^{3}$ – $5^{5}$ – … – $5^{99}$
=> 24B = $5^{101}$ -5
=> B = $\frac{5^{101}-5}{24}$
+) H = (1+2+3+ … +100)($1^{2}$+$2^{2}$+$3^{2}$ +…+$100^{2}$)(65.111-13.5.111)
H = (1+2+3+ … +100)($1^{2}$+$2^{2}$+$3^{2}$ +…+$100^{2}$)(111.65-111.65)
H = (1+2+3+ … +100)($1^{2}$+$2^{2}$+$3^{2}$ +…+$100^{2}$).0
H = 0
$E=2+2^2+2^3+…+2^{100}$
$⇒2E=2^2+2^3+2^4+…+2^{101}$
$⇒2E-E=(2^2+2^3+2^4+…+2^{101})-(2+2^2+2^3+…+2^{100})$
$⇒E=2^{101}-2$
Vậy $E=2^{101}-2$
$F=1+3^1+3^2+…+3^{100}$
$⇒3F=3^1+3^2+3^3+…+3^{101}$
$⇒3F-F=(3^1+3^2+3^3+…+3^{101})-(1+3^1+3^2+…+3^{100})$
$⇒2F=3^{101}-1$
$⇒F=\frac{3^{101}-1}{2}$
Vậy $F=\frac{3^{101}-1}{2}$
$G=5+5^3+5^5+…+5^{99}$
$⇒25G=5^3+5^5+5^7+…+5^{101}$
$⇒25G-G=(5^3+5^5+5^7+…+5^{101})-(5+5^3+5^5+…+5^{99})$
$⇒24G=5^{101}-5$
$⇒G=\frac{5^{101}-5}{24}$
Vậy $G=\frac{5^{101}-5}{24}$
$H=(1+2+3+…+100).(1^2+2^2+3^2+…+100^2).(65.111-13.15.37)$
$=(1+2+3+…+100).(1^2+2^2+3^2+…+100^2).[65.111-(13.5).(3.37)]$
$=(1+2+3+…+100).(1^2+2^2+3^2+…+100^2).(65.111-65.111)$
$=(1+2+3+…+100).(1^2+2^2+3^2+…+100^2).0$
$=0$
Vậy $H=0$.