【高校数学B】1-2-1 和の記号｜問題集

\(\displaystyle =\frac{1(5^n-1)}{5-1}\)
\(\displaystyle =\frac{1}{4}(5^n-1)\)

\(\displaystyle =\frac{3(3^{n-1}-1)}{3-1}\)
\(\displaystyle =\frac{3}{2}(3^{n-1}-1)\)

\(\displaystyle =4・\frac{1}{2}n(n+1)-5n\)
\(=2n^2+2n-5n\)
\(=n(2n-3)\)

\(\displaystyle =\frac{1}{6}n(n+1)(2n+1)-3・\frac{1}{2}n(n+1)+2n\)
\(\displaystyle =\frac{1}{6}n(2n^2+3n+1-9n-9+12)\)
\(\displaystyle =\frac{1}{6}n(2n^2-6n+4)\)
\(\displaystyle =\frac{1}{3}n(n^2-3n+2)\)
\(\displaystyle =\frac{1}{3}n(n-1)(n-2)\)

\(\displaystyle =\sum_{k=1}^{n}(k^3+k)\)
\(\displaystyle =\left\{\frac{1}{2}n(n+1)\right\}^2+\frac{1}{2}n(n+1)\)
\(\displaystyle =\frac{1}{4}n(n+1)\{n(n+1)+2\}\)
\(\displaystyle =\frac{1}{4}n(n+1)(n^2+n+2)\)

\(\displaystyle =2・\frac{1}{2}n(n-1)+(n-1)\)
\(\displaystyle =n(n-1)+(n-1)\)
\(\displaystyle =(n+1)(n-1)\)

\(\displaystyle =2・\frac{1}{2}・2n(2n+1)+3・2n\)
\(\displaystyle =2n(2n+4)\)
\(\displaystyle =4n(n+2)\)

\(\displaystyle =\frac{1}{2}n(n+1)+2n\)
\(\displaystyle =\frac{1}{2}n(n+1+4)\)
\(\displaystyle =\frac{1}{2}n(n+5)\)

\(\displaystyle =\frac{\frac{1}{2}(1-(\frac{1}{2})^n)}{1-\frac{1}{2}}\)
\(\displaystyle =1-\left(\frac{1}{2}\right)^n\)

\(=2・15\)
\(=30\)

\(\displaystyle =\frac{1}{2}・20・21\)
\(\displaystyle =210\)

\(\displaystyle \sum_{k=1}^{n}(k+1)^2\)
\(\displaystyle =\sum_{k=1}^{n}(k^2+2k+1)\)
\(\displaystyle =\frac{1}{6}n(n+1)(2n+1)+2・\frac{1}{2}n(n+1)+n\)
\(\displaystyle =\frac{1}{6}n(2n^2+3n+1+6n+6+6)\)
\(\displaystyle =\frac{1}{6}n(2n^2+9n+13)\)
よって、
\(\displaystyle \sum_{k=1}^{10}(k+1)^2\)
\(\displaystyle =\frac{1}{6}10(2・10^2+9・10+13)\)
\(\displaystyle =505\)

\(\displaystyle =\sum_{k=1}^{12}k^2\)
\(\displaystyle =\frac{1}{6}・12(12+1)(2・12+1)\)
\(\displaystyle =650\)

一般項\(a_n=3n-1\)
\(\displaystyle \sum_{k=1}^{n}(3k-1)\)
\(\displaystyle =3・\frac{1}{2}n(n+1)-n\)
\(\displaystyle =\frac{1}{2}n(3n+1)\)

一般項\(a_n=n^3\)
\(\displaystyle \sum_{k=1}^{n}k^3\)
\(\displaystyle =\left\{\frac{1}{2}n(n+1)\right\}^2\)
\(\displaystyle =\frac{1}{4}n^2(n+1)^2\)