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A sequence is defined recursively by f(1)=16 and f(n)= f(n-1)+2n. Find f(4)

lord [1]

Hi,

f(1)=16
f(2)=f(1)+2*2=16+4=20
f(3)=f(2)+2*3=20+6=26
f(4)=f(3)+2*4=26+8=34

8 0

3 years ago

How can i prove this property to be true for all values of n, using mathematical induction.

chubhunter [2.5K]

Proof -

So, in the first part we'll verify by taking n = 1.

$\implies \: 1 = {1}^{2} = \frac{1(1 + 1)(2 + 1)}{6}$

$\implies{ \frac{1(2)(3)}{6} }$

$\implies{ 1}$

Therefore, it is true for the first part.

In the second part we will assume that,

$\: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} }$

and we will prove that,

$\sf{ \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}$

$\: {{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 2) (2k + 3)}{6}}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} + \frac{(k + 1) ^{2} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2 +k+6k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6}$

Henceforth, by using the principle of mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n.

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8 0

2 years ago

two golfers completed one round of golf. the first golfer had a score of +6 and the second golfer had a score of -3. how many mo

pav-90 [236]

9. One is 6 over par, the other is three under, so subtract first-second = 6-(-3) = 9

4 0

4 years ago

A homeowner is installing a swimming pool in his backyard. he want its length to be 4 feet longer than its width. then he wants

marishachu [46]

1. You have that:

- The homeowner want the length of the swimming pool to be 4 feet longer than its width.
- He wants to surround it with a concrete walkway 3 feet wide.
- He can only afford 300 square feet of concrete for the walkway.

2. Therefore, the tota area is:

At=L2xW2

L2 is the lenght of the walkway (L2=L1+3+3⇒L2=(W1+4+6)⇒L2=W1+10).
W2 is the width of the walkway (W1+3+3⇒W2=W1+6)

3. The area of the walkway is:

A2=At-A1

A2=300 ft²

4. Therefore, you have that the width of the swimming pool is:

A2=(W1+10)(W1+6)-(W1+4)(W1)
300=(W1²+6W1+10W1+60)-(W1²+4W1)
W1²+16W1+60-W1²-4W1-300=0
12W1-240=0
W1=240/12
W1=20 ft

5. And the length is:

L1=W1+4
L1=20+4
L1=24 ft

8 0

3 years ago

If you spend 7 hours a day sleeping what fraction of the day are you asleep?

Viktor [21]

Answer:

7/24

Step-by-step explanation:

3 0

4 years ago

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