Use mathematical induction to prove that the following statement is true for every positive integer n
1 answer:
<h3>to prove
</h3><h3>
</h3><h3>8
</h3><h3>+
</h3><h3>16
</h3><h3>+
</h3><h3>24
</h3><h3>+
</h3><h3>...
</h3><h3>+
</h3><h3>8
</h3><h3>n
</h3><h3>=
</h3><h3>4
</h3><h3>n
</h3><h3>(
</h3><h3>n
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>−
</h3><h3>−
</h3><h3>−
</h3><h3>−
</h3><h3>(
</h3><h3>*
</h3><h3>)
</h3><h3>
</h3><h3> let </h3><h3>T
</h3><h3>n
</h3><h3>=
</h3><h3>4
</h3><h3>n
</h3><h3>(
</h3><h3>n
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>
</h3><h3>(1) verify for </h3><h3>n
</h3><h3>=
</h3><h3>1
</h3><h3>
</h3><h3>L
</h3><h3>H
</h3><h3>S
</h3><h3>=
</h3><h3>8
</h3><h3>
</h3><h3>R
</h3><h3>H
</h3><h3>S
</h3><h3>=
</h3><h3>4
</h3><h3>×
</h3><h3>1
</h3><h3>(
</h3><h3>1
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>=
</h3><h3>4
</h3><h3>×
</h3><h3>2
</h3><h3>=
</h3><h3>8
</h3><h3>
</h3><h3>∴
</h3><h3>true for </h3><h3>n
</h3><h3>=
</h3><h3>1
</h3><h3>
</h3><h3># to show
</h3><h3>
</h3><h3>T
</h3><h3>k
</h3><h3>⇒
</h3><h3>T
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>
</h3><h3>assume true for </h3><h3>T
</h3><h3>k
</h3><h3>=
</h3><h3>4
</h3><h3>k
</h3><h3>(
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>
</h3><h3>need to show
</h3><h3>
</h3><h3>T
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>=
</h3><h3>4
</h3><h3>(
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>(
</h3><h3>k
</h3><h3>+
</h3><h3>2
</h3><h3>)
</h3><h3>
</h3><h3>add next term to to both sides of </h3><h3>(
</h3><h3>*
</h3><h3>)
</h3><h3>
</h3><h3>8
</h3><h3>+
</h3><h3>16
</h3><h3>+
</h3><h3>24
</h3><h3>+
</h3><h3>...
</h3><h3>+
</h3><h3>8
</h3><h3>k
</h3><h3>+
</h3><h3>8
</h3><h3>(
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>=
</h3><h3>4
</h3><h3>k
</h3><h3>(
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>+
</h3><h3>8
</h3><h3>(
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>
</h3><h3>∴
</h3><h3>T
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>=
</h3><h3>4
</h3><h3>k
</h3><h3>(
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>+
</h3><h3>8
</h3><h3>(
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>
</h3><h3>=
</h3><h3>4
</h3><h3>(
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>)
</h3><h3>[
</h3><h3>k
</h3><h3>+
</h3><h3>2
</h3><h3>]
</h3><h3>=
</h3><h3>T
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>
</h3><h3>i
</h3><h3>.
</h3><h3>e
</h3><h3>.
</h3><h3>T
</h3><h3>k
</h3><h3>⇒
</h3><h3>T
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3> as required
</h3><h3>
</h3><h3>#(3) conclusion
</h3><h3>
</h3><h3>statement true for </h3><h3>T
</h3><h3>1
</h3><h3>
</h3><h3>∵
</h3><h3>T
</h3><h3>k
</h3><h3>⇒
</h3><h3>T
</h3><h3>k
</h3><h3>+
</h3><h3>1
</h3><h3>
</h3><h3>T
</h3><h3>1
</h3><h3>⇒
</h3><h3>T
</h3><h3>2
</h3><h3>⇒
</h3><h3>T
</h3><h3>3
</h3><h3>⇒
</h3><h3>...
</h3><h3>∀
</h3><h3>n
</h3><h3>∈
</h3><h3>N </h3>
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Answer: 3/5
Step-by-step explanation: The fourth quadratic in the list is the square of a binomial:
25x^2 - 30x + 9 = (5x - 3)^2
Therefore, the repeated root of this quadratic is the solution to 5x - 3=0, which is 3/5
Step-by-step explanation:
27/4 × 17/3
9/4 × 17/1
153/4
Answer:
-2
Step-by-step explanation:
(-5) + (+3)
= -5+3
= -2
please mark as brainliest
Answer:
nothing
Step-by-step explanation:
Loan payments are <em>linear in the loan amount</em> for a given rate and period, so the payments for loans of $2000 and $1000 sum to the amount of payments for a loan of $3000.
The only possible savings (or cost) might come from rounding to the nearest cent. (In any event, the final payment on each loan should make up for any differences due to rounding.)
Answer:
here,
principle(p)=£8000
time(t)=5 years
rate(r)=10%
now,
Interest=PTR/100
=8000 x 5 x 10/100
=£4000
Step-by-step explanation:
