Can someone tell me how 15n-n^2+16 can become 17(n+1)-(n+1)^2? This is induction method?
回答 (2)
2018-06-04 12:11 am
L.H.S.
= 15n - n² + 16
= (16n - n) - n² + 16
= 16n - n² + 16 - n
= (16n - n²) + (16 - n)
= n(16 - n) + (16 - n)
= (16 - n) (n + 1)
= (17 - n - 1) (n + 1)
= [17 - (n + 1)] (n + 1)
= 17 (n + 1) - (n + 1)²
= R.H.S.
Hence, 15n - n² + 16 = 17(n + 1) - (n + 1)²
= 15n - n² + 16
= (16n - n) - n² + 16
= 16n - n² + 16 - n
= (16n - n²) + (16 - n)
= n(16 - n) + (16 - n)
= (16 - n) (n + 1)
= (17 - n - 1) (n + 1)
= [17 - (n + 1)] (n + 1)
= 17 (n + 1) - (n + 1)²
= R.H.S.
Hence, 15n - n² + 16 = 17(n + 1) - (n + 1)²
2018-06-04 3:14 am
It is very easy to show that
17(n+1)-(n+1)² = 17n + 17 - (n² + 2n + 1)
. . . . . . . . . . . . = -n² + 15n + 16
. . . . . . . . . . . . = 15n - n² + 16
So the two expressions are identical. However, this has nothing at all to do with inductance.
So I have no idea what you mean by "become" in this instance. Have you told us the entire question ?
17(n+1)-(n+1)² = 17n + 17 - (n² + 2n + 1)
. . . . . . . . . . . . = -n² + 15n + 16
. . . . . . . . . . . . = 15n - n² + 16
So the two expressions are identical. However, this has nothing at all to do with inductance.
So I have no idea what you mean by "become" in this instance. Have you told us the entire question ?
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