1/(2*3)+1/(3*4)+1/(4*5)+…+1/(2016*2017)?

Hitunglah nilai \displaystyle \sum\limits_{n=1}^{2015}{\frac{1}{\left( n+1 \right)\left( n+2 \right)}}.

Jawaban:

\sum\limits_{n=1}^{2015}{\frac{1}{\left( n+1 \right)\left( n+2 \right)}}=\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\cdots +\frac{1}{2015\cdot 2016}+\frac{1}{2016\cdot 2017}

Setiap pecahan ini bisa kita pecah menjadi seperti berikut:

\displaystyle \begin{array}{l}\frac{1}{2\cdot 3}=\frac{3-2}{2\cdot 3}=\frac{3}{2\cdot 3}-\frac{2}{2\cdot 3}=\frac{1}{2}-\frac{1}{3}\\\frac{1}{3\cdot 4}=\frac{4-3}{3\cdot 4}=\frac{4}{3\cdot 4}-\frac{3}{3\cdot 4}=\frac{1}{3}-\frac{1}{4}\\\frac{1}{4\cdot 5}=\frac{5-4}{4\cdot 5}=\frac{5}{4\cdot 5}-\frac{4}{4\cdot 5}=\frac{1}{4}-\frac{1}{5}\\\frac{1}{2015\cdot 2016}=\frac{1}{2015}-\frac{1}{2016}\\\frac{1}{2016\cdot 2017}=\frac{1}{2016}-\frac{1}{2017}\end{array}

Kembali ke persamaan awal, maka diperoleh:

\displaystyle \sum\limits_{n=1}^{2015}{\frac{1}{\left( n+1 \right)\left( n+2 \right)}}=\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\left( \frac{1}{4}-\frac{1}{5} \right)+\cdots +\left( \frac{1}{2015}-\frac{1}{2016} \right)+\left( \frac{1}{2016}-\frac{1}{2017} \right)

Selanjutnya, tinggal mengubah susunan penjumlahan untuk perhitungan lebih mudah. Semuanya menghasilkan nol, menyisakan dua angka pertama dan terakhir.

\displaystyle \begin{array}{l}\sum\limits_{n=1}^{2015}{\frac{1}{\left( n+1 \right)\left( n+2 \right)}}=\frac{1}{2}+\left( -\frac{1}{3}+\frac{1}{3} \right)+\left( -\frac{1}{4}+\frac{1}{4} \right)+\cdots +\left( -\frac{1}{2016}+\frac{1}{2016} \right)-\frac{1}{2017}\\\sum\limits_{n=1}^{2015}{\frac{1}{\left( n+1 \right)\left( n+2 \right)}}=\frac{1}{2}+\left( 0 \right)+\left( 0 \right)+\cdots +\left( 0 \right)-\frac{1}{2017}\end{array}

Sehingga diperoleh:

\displaystyle \sum\limits_{n=1}^{2015}{\frac{1}{\left( n+1 \right)\left( n+2 \right)}}=\frac{1}{2}-\frac{1}{2017}

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