$*Tính tổng: S=1^{2}+3^{2}+5^{2}+...+(2n-1)^{2}$
$Giải:$
$Đặt: S(2n-1)=1^{2}+3^{2}+5^{2}+...+(2n-1)^{2}$
$\text{S(2n)=$2^{2}$+$4^{2}$+$6^{2}$+...+$(2n)^{2}$=$\frac{4n(n+1)(2n+1)}{6}$}$
$\text{⇒S(2n−1)+S(2n)=$1^{2}$+$2^{2}$+$3^{2}$+$4^{2}$+...+$(2n-1)^{2}$}$
$\text{+$(2n)^{2}$(*)}$
$Ta có:$$1^{2}$+$2^{2}$+...$n^{2}$=$\frac{n(n+1)(2n+1)}{6}$
$\text{⇒(∗)⇔S(2n−1)+S(2n)=$\frac{(2n)(2n+1)(4n+1)}{6}$}$
$\text{⇔S(2n−1)=$\frac{(2n)(2n+1)(4n+1)}{6}$$\frac{4n(n+1)(2n+1)}{6}$}$
$\text{⇔S(2n-1)=$\frac{(2n)(2n+1)}{6}$(4n+1-2n-2)=$\frac{n(2n−1)(2n+1)}{3}$ }$
⇒$1^{2}$ + $2^{2}$ + $5^{2}$ + ... $(2n-1)^{2}$=$\frac{n(2n-1)(2n+1)}{3}$
$*Tính tổng: K=1^{3}+3^{3}+5^{3}+...+(2n-1)^{3}$
$Giải:$
$Ta có: 1^{3}+3^{3}+5^{3}+...+(2n-1)^{3}$$=1^{3}+2^{3}+3^{3}+...+(2n)^{3}$-[$2^{3}$+$4^{3}$+$6^{3}$+...+$(2n)^{3}]$
$\text{Áp dụng kết quả trên ta có: $1^{3}$+$2^{3}$+$3^{3}$+...+$(2n)^{3}$=$n^{2}$$(2n+1)^{2}$}$
$\text{Vậy K=$1^{3}$+$3^{3}$+$5^{3}$+...+$(2n-1)^{3}$ }$=$n^{2}$$(2n+1)^{2}$-$2n^{2}$$(n+1)^{2}$=$2n^{4}$-$n^{2}$
#Viết latex mệt quá.Tự là câu D=$2^{2}$+$4^{2}$+$6^{2}$ + ... + $(2n-2)^{2}$ nha!
#Học tốt! ^^ Xin ctrlhn ạ!
trinhthuy1987-Hoidap247.
