Question

Here are the first 5 terms of a sequence 3,8,17,30,47 find an expression, in terms of n for the nth term of this sequence

Answer 1

The first differences are 5, 9, 13, 17, and so the second differences are all 4.

Halving 4 gives 2, so the first term of the sequence is 2n^2.

Subtracting 2n^2 from the sequences gives 1,0,-1-2,-3 which has the nth term -n+2.

Therefore, the formula for this sequence is 2n^2 -n +2.

Answer 2

2n^2 - n +2

Explanation:

3,8,17,30,47

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this is the final form I got after playing for a while:

• 1. 3= 4*1 + 1 - 2= 4*(1+0) + 1 -2 = 4*(1 + (1-2)(1-1)/2) + 1- 2
• 2. 8= 4*2 + 0 = 4*(2+0)+0= 4*(2+0) + 2-2= 4*(2+ (2-2)(2-1)/2) + 2-2
• 3. 17= 4*4 + 1= 4*(3+1) +1= 4*(3+1) + 3-2= 4*(3+ (3-2)(3-1)/2) + 3-2
• 4. 30= 4*7 +2= 4*(4+3)+2= 4*(4+3) + 4-2= 4*(4+ (4-2)(4-1)/2) + 4-2
• 5. 47= 4*11+ 3= 4*(5+6)+3 = 4*(5+6) + 5-2= 4*(5+(5-2)(5-1)/2) + 5-2

.....

• n. an= 4*(n+(n-2)(n-1)/2)+ n- 2= 4n + 2(n^2 -3n+2) + n-2=

• =4n +2n^2 - 6n +4 + n - 2= 2n^2 - n +2

• an= 2n^2 - n +2

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tested in excel and works:

1 2 3 4 5 6 7 8 9 10

3 8 17 30 47 68 93 122 155 192