Soal dan Pembahasan OSN Matematika SMP Tingkat Provinsi 2012 Nomor 3

Soal:

Jika 1 + (1/4) + (1/9) + (1/16) + (1/25) + ..... = a, maka (1/9) + (1/25) + (1/49) + ..... = .....

Pembahasan:
1 + (1/4) + (1/9) + (1/16) + (1/25) + (1/36) + (1/49) + ..... = a

Kemudian:

==> (1/4) + (1/9) + (1/16) + (1/25) + (1/36) + (1/49) +..... = a – 1 

==> (1/9) + (1/25) + (1/49) +..... = a – 1– [(1/4) + (1/16) + (1/36)+ .....]

==> (1/9) + (1/25) + (1/49) +..... = a – 1– (1/4)[(1) + (1/4) + (1/9)+ .....]

==> (1/9) + (1/25) + (1/49) +..... = a – 1– (1/4)(a)

==> (1/9) + (1/25) + (1/49) +..... = (3/4)a – 1

Jadi, (1/9) + (1/25) + (1/49) +..... = (3/4)a – 1

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Pelajari juga: [Nomor 1], [Nomor 2]