Mengubah bentuk Batas Bawah menjadi suku Ke-Un
Untuk menyelesaikan soal-soal dibawah ini
Bentuk
 "\sum_{n=2}^{11}\left ( 3n-4 \right )")
di ubah menjadi
 "\sum_{n=7}^{11}\left ( U_{n} \right )")
maka nilai Un adalah
Penyelesaian
=\sum_{n=2}^{6}\left&space;(&space;3n-4&space;\right&space;)+&space;\sum_{n=7}^{11}\left&space;(&space;3n-4&space;\right&space;) "\sum_{n=2}^{11}\left ( 3n-4 \right )=\sum_{n=2}^{6}\left ( 3n-4 \right )+ \sum_{n=7}^{11}\left ( 3n-4 \right )")
=\sum_{n=2+5}^{6+5}\left&space;(&space;3(n-5)-4&space;\right&space;)+&space;\sum_{n=7}^{11}\left&space;(&space;3n-4&space;\right&space;) "\sum_{n=2}^{11}\left ( 3n-4 \right )=\sum_{n=2+5}^{6+5}\left ( 3(n-5)-4 \right )+ \sum_{n=7}^{11}\left ( 3n-4 \right )")
=\sum_{n=7}^{11}\left&space;(&space;3n-15-4&space;\right&space;)+&space;\sum_{n=7}^{11}\left&space;(&space;3n-4&space;\right&space;) "\sum_{n=2}^{11}\left ( 3n-4 \right )=\sum_{n=7}^{11}\left ( 3n-15-4 \right )+ \sum_{n=7}^{11}\left ( 3n-4 \right )")
=\sum_{n=7}^{11}\left&space;(&space;3n-19)+(3n-4) "\sum_{n=2}^{11}\left ( 3n-4 \right )=\sum_{n=7}^{11}\left ( 3n-19)+(3n-4)")
=\sum_{n=7}^{11}\left&space;(&space;6n-23) "\sum_{n=2}^{11}\left ( 3n-4 \right )=\sum_{n=7}^{11}\left ( 6n-23)")
Jadi jawabanya adalah
 "\sum_{n=7}^{11}\left ( 6n-23)")
Semoga Bermanfaat
Bentuk
di ubah menjadi
maka nilai Un adalah
Penyelesaian
Jadi jawabanya adalah
Semoga Bermanfaat
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