Using the principle of mathematical induction prove that for all n belongs to N: 6 power"n+2" + 7 power"2n+1" is divisible by 43
Asked by Theodosia Lourdes
| 8th May, 2011,
03:53: AM
Expert Answer:
Let P(n) be the statement ( 6 (n+2)+ 7(2n+1)) = 43x.
- Basis of Induction
Since ( 6(1+2)+ 7(2+1)) = 63 + 73 = 559 = 43 X 13,
the formula is true for n = 1.
- Inductive Hypothesis
Assume that P(n) is true for n = k, that is 6(k+1)+ 7(2k+1)) = 43x
for some integer x.
- Inductive Step
Now show that the formula is true for n = k + 1.
Observe that
P(k+1) = 6(k+1+2)+ 7(2(k+1)+1)
= 6(k+3)+ 7(2k+3)
= 6 X 6(k+2)+ 72 X 7(2k+1)
= 6 X 6(k+2)+(6+43) X 7(2k+1)
= 6 X 6(k+2)+ 6 X 7(2k+1)+ 43 X 7(2k+1)
= 6 X ( 6(k+2)+ 7(2k+1) )+ 43 X 7(2k+1)
= 6 X ( P(k) ) + 43 X 7(2k+1)
Since each component of this sum is divisible by 43 so is the entire sum and the formula holds for k + 1.
Since ( 6(1+2)+ 7(2+1)) = 63 + 73 = 559 = 43 X 13,
the formula is true for n = 1.
Assume that P(n) is true for n = k, that is 6(k+1)+ 7(2k+1)) = 43x
for some integer x.
Now show that the formula is true for n = k + 1.
Observe that
P(k+1) = 6(k+1+2)+ 7(2(k+1)+1)
= 6(k+3)+ 7(2k+3)
= 6 X 6(k+2)+ 72 X 7(2k+1)
= 6 X 6(k+2)+(6+43) X 7(2k+1)
= 6 X 6(k+2)+ 6 X 7(2k+1)+ 43 X 7(2k+1)
= 6 X ( 6(k+2)+ 7(2k+1) )+ 43 X 7(2k+1)
= 6 X ( P(k) ) + 43 X 7(2k+1)
Since each component of this sum is divisible by 43 so is the entire sum and the formula holds for k + 1.
Answered by
| 8th May, 2011,
10:27: AM
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