<p>Jawaban : n+1</p><p>&nbsp;</p><p>Ingat!</p><p>Pola teleskopik adalah pola yang saling menghilangkan</p><p>&nbsp;</p><p>Dik:&nbsp;</p><h1>(1+(1/1)).(1+(1/2)).(1+(1/3)).(1+(1/4)). ... .(1+(1/n)).</h1><p>agar dapat saling menghilangkan, kita harus ubah bentuk bilangan diatas</p><p>&nbsp;(1+(1/1)).(1+(1/2)).(1+(1/3)).(1+(1/4)). ... .(1+(1/n)).</p><p>=(1+(1)).(1+(1/2)).(1+(1/3)).(1+(1/4)). ... .(1+(1/(n-1))) . (1+(1/n)).</p><p>=2.(3/2).(4/3).(5/4). ... .(1+(1/(n-1))) . (1+(1/n)).</p><p>&nbsp;</p><p>kita ubah suku terakhir (1+(1/n))</p><p>(1+(1/n))=(n/n+1/n)</p><p>=(n+1)/n</p><p>Kita ubah suku kedua terakhir (1+(1/(n-1)))</p><p>(1+(1/(n-1)))=((n-1)/(n-1)+(1/(n-1)))</p><p>=((n-1)+1)/(n-1)</p><p>=n/(n-1)</p><p>&nbsp;</p><p>Kembali ke pola tadi</p><p>=2.(3/2).(4/3).(5/4). ... .(1+(1/(n-1))) . (1+(1/n)).</p><p>=(2/1).(3/2).(4/3).(5/4). ... .n/(n-1) . (n+1)/n.</p><p>&nbsp;</p><p>setelah saling menghilangkan, didapat</p><p>(n+1)/1=n+1</p><p>&nbsp;</p><p>Jadi, formula yang memenuhi setiap pola tersebut adalah n+1</p>