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Proof by mathematical induction

Let n≥ 3 be...

Question

Let $n \geq 3$ be an integer. For a permutation $σ = (a_{1}, a_{2}, . . . . . . a_{n})$ of $(1, 2, . . . ., n)$ we let of $σ (x) = a_{n} x^{n - 1} + a^{n - 1} + . . . . + a_{2} x + a_{1}$ . Let $S_{σ}$ be the sum of the roots of $f_{σ} (x) = 0$ and let S denote the sum over all permutation $σ$ of $(1, 2, . . . ., n)$ of the number $S_{σ}$ . Then.

A

S < - n!

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B

- n! < S < 0

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C

0 < S < n!

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D

n! < S

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Solution

The correct option is A $S < - n!$
its difficult to go for general n so put n=3 now sum of roots is always - $(a_{n} - 1) / a_{n}$ so there are six combination $1 / 2, 1 / 3, 2 / 1, 3 / 1, 2 / 3, 3 / 2$ summing them up we get $S$ = - $8$ which is $< 3!$ so $A$ is correct as in our case n=3

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Similar questions

Q. Let n be an odd integer greater than 1 and let

c_{1}, c_{2}, \dots, c_{n}

be integers. For each permutation

a = (a_{1}, a_{2}, \dots, a_{n})

(1, 2, \dots, n)

S (a) = n \sum i = 1 c_{i} a_{i} .

Prove that there exist permutations

a \neq b

{1, 2, \dots, n}

such that n! is a divisor of S(a)-S(b).

S_{n}

denotes the sum of first

n

terms of an A.P . If

S_{3 n} = 5 S_{n}

S_{2 n} : S_{n}

.

α, β

denote the cube roots of unity other than

1

α β

s = \sum_{n = 0}^{2} {(- 1)}^{n} {(\frac{α}{β})}^{n}

. Then the value of

s

n \geq 1

be a positive integer. Suppose, between two distinct positive real numbers

k

l, n

arithmetic means

a_{1}, a_{2}, . . ., a_{n}

n

h_{1}, h_{2}, . . . ., h_{n}

are inserted. For

i = 1, 2, . . . ., n

j = 1, 2, . . . ., n

p (i, j) = a_{i} h_{j}

n

be a positive integer. Call a non-empty subset

S

{1, 2, . . ., n}

good, if the arithmetic mean of the elements of

S

, is also an integer. Further let

t_{n}

denote the number of good subsets of

{1, 2, . . ., n}

t_{n}

n

are both odd or both even.

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