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Question Number 147791 by Tawa11 last updated on 23/Jul/21

If a + b + c + d = 1

a^{2} + b^{2} + c^{2} + d^{2} = 2

a^{3} + b^{3} + c^{3} + d^{3} = 3

a^{4} + b^{4} + c^{4} + d^{4} = 4

Evaluate :

a^{6} + b^{6} + c^{6} + d^{6}

Commented by prakash jain last updated on 25/Jul/21

g (k) = \frac{1}{1 - k a} = 1 + k a + k^{2} a^{2} + . .

f (k) = \frac{1}{1 - k a} + \frac{1}{1 - k b} + \frac{1}{1 - k c} + \frac{1}{1 - k d}

f (k) = \frac{4 + A_{1} k + A_{2} k^{2} + A_{3} k^{3}}{1 + B_{1} k + B_{2} k^{2} + B_{3} k^{3} + B_{4} k^{2}}

A_{1} = - 3 (a + b + c + d)

B_{1} = - (a + b + c + d) \Rightarrow A_{1} = 3 B_{1}

A_{2} = 2 (a b + b c + a d + b c + b d + c d) = 2 B_{2}

A_{3} = - (a b c + a b d + a c d + b c d) = B_{3}

f (k) = 4 + k (a + b + c + d) + k^{2} (a^{2} + b^{2} + c^{2} + d^{2}) . .

Let us denote a^{n} + b^{n} + c^{n} + d^{n} = u_{n}

f (k) = 4 + {k u}_{1} + k^{2} u_{2} + k^{3} u_{3} + k^{4} u_{4} + \dots

4 + A_{1} k + A_{2} k^{2} + A_{3} k^{3} =

(1 + B_{1} k + B_{2} k^{2} + B_{3} k^{3}) (4 + {k u}_{1} + k^{2} u_{2} + k^{3} u_{3} + k^{4} u_{4} + k^{5} u_{5} + k^{6} u 6 + \dots)

Given u_{1} = 1, u_{2} = 2, u_{3} = 3, u_{4} = 4

C o m p a r i n g c o e f f i c i e n t s

A_{1} = u_{1} + 4 B_{1} \Rightarrow 3 B_{1} = 1 + 4 B_{1}

\Rightarrow B_{1} = - 1, A_{1} = - 3

A_{2} = u_{2} + u_{1} B_{1} + 4 B_{2} = 2 - 1 + 4 B_{2}

\Rightarrow 2 B_{2} = 1 + 4 B_{2} \Rightarrow B_{2} = \frac{- 1}{2}, A_{2} = - 1

A_{3} = u_{3} + B_{1} u_{2} + B_{2} u_{1} + 4 B_{3}

\Rightarrow B_{3} = 3 - 2 - \frac{1}{2} + 4 B_{3} \Rightarrow B_{3} = A_{3} = - \frac{1}{6}

0 = u_{4} + B_{1} u_{3} + B_{2} u_{2} + B_{3} u_{1} + 4 B_{4}

\Rightarrow 0 = 4 - 3 - \frac{1}{2} \times 2 - \frac{1}{6} + 4 B_{4} \Rightarrow B_{4} = \frac{1}{24}

0 = u_{5} + B_{1} u_{4} + B_{2} u_{3} + B_{3} u_{2} + B_{4} u_{1}

u_{1} . . u_{4}, B_{1} \dots B_{4} are known u_{5} can be

calculated

similary

0 = u_{6} + B_{1} u_{5} + B_{2} u_{4} + B_{3} u_{3} + B_{4} u_{2}

u_{1} \dots u_{5} are known and you can

calculate u_{6} = a^{6} + b^{6} + c^{6} + d^{6}

Commented by prakash jain last updated on 25/Jul/21

u_{5} - 4 - \frac{3}{2} - \frac{1}{3} + \frac{1}{24} = 0

u_{5} = \frac{96 + 36 + 8 - 1}{24} = \frac{139}{24}

u_{6} - \frac{139}{24} - \frac{1}{2} \times 4 - \frac{1}{6} \times 3 + \frac{1}{24} \times 2

u_{6} = \frac{139}{24} + 2 + \frac{1}{2} - \frac{1}{12} = \frac{139 + 48 + 12 - 2}{24}

= \frac{197}{24}

a^{6} + b^{6} + c^{6} + d^{6} = \frac{197}{24}

I {haven}^{'} t checked numerical calculation

thoroughly so please recheck these .

You can use this method to find

any u_{n} = a^{n} + b^{n} + c^{n} + d^{n}

Commented by Tawa11 last updated on 25/Jul/21

Thanks sir . I appreciate . God bless you .

Answered by Rasheed.Sindhi last updated on 23/Jul/21

L e t b = pa, c = qa, d = ra

(i) \Rightarrow a (1 + p + q + r) = 1

\Rightarrow p + q + r = 1 / a - 1 = (1 - a) / a \dots A

(ii) \Rightarrow a^{2} (1 + p^{2} + q^{2} + r^{2}) = 2

\Rightarrow p^{2} + q^{2} + r^{2} = 2 / a^{2} - 1 = (2 - a^{2}) / a^{2} \dots B

(iii) \Rightarrow p^{3} + q^{3} + r^{3} = (3 - a^{3}) / a^{3} \dots C

(iv) \Rightarrow p^{4} + q^{4} + r^{4} = (4 - a^{4}) / a^{4} \dots D

A \Rightarrow {(p + q + r)}^{2} = {(1 - a) / a}^{2}

p^{2} + q^{2} + r^{2} + 2 (pq + qr + rp) = {(\frac{1 - a}{a})}^{2}

\frac{2 - a^{2}}{a^{2}} + 2 (pq + qr + rp) = {(\frac{1 - a}{a})}^{2}

2 (pq + qr + rp) = {(\frac{1 - a}{a})}^{2} - \frac{2 - a^{2}}{a^{2}}

pq + qr + rp = \frac{{(1 - a)}^{2} - 2 + a^{2}}{{2 a}^{2}}

C \Rightarrow p^{3} + q^{3} + r^{3} = \frac{3 - a^{3}}{a^{3}}

p^{3} + q^{3} + r^{3} - 3 pqr = \frac{3 - a^{3}}{a^{3}} - 3 pqr

(p + q + r) (p^{2} + q^{2} + r^{2} - (pq + qr + rp))

= \frac{3 - a^{3}}{a^{3}} - 3 pqr

3 pqr = \frac{3 - a^{3}}{a^{3}} - (\frac{1 - a}{a}) (\frac{2 - a^{2}}{a^{2}} - \frac{{(1 - a)}^{2} - 2 + a^{2}}{{2 a}^{2}})

pqr = \frac{3 - a^{3}}{{3 a}^{3}} - (\frac{1 - a}{3 a}) (\frac{2 - a^{2}}{a^{2}} - \frac{{(1 - a)}^{2} - 2 + a^{2}}{{2 a}^{2}})

Commented by Tawa11 last updated on 23/Jul/21

Thanks sir . God bless you .

Answered by mr W last updated on 23/Jul/21

p_{k} = a^{k} + b^{k} + c^{k} + d^{k}

e_{1} = p_{1} = 1

2 e_{2} = e_{1} p_{1} - p_{2} = 1 - 2 = - 1 \Rightarrow e_{2} = - \frac{1}{2}

3 e_{3} = e_{2} p_{1} - e_{1} p_{2} + p_{3} = - \frac{1}{2} - 2 + 3 = \frac{1}{2} \Rightarrow e_{3} = \frac{1}{6}

4 e_{4} = e_{3} p_{1} - e_{2} p_{2} + e_{1} p_{3} - p_{4} = \frac{1}{6} + 1 + 3 - 4 = \frac{1}{6} \Rightarrow e_{4} = \frac{1}{24}

0 = e_{4} p_{1} - e_{3} p_{2} + e_{2} p_{3} - e_{1} p_{4} + p_{5}

\Rightarrow p_{5} = - \frac{1}{24} + \frac{1}{3} + \frac{3}{2} + 4 = \frac{139}{24}

0 = - e_{4} p_{2} + e_{3} p_{3} - e_{2} p_{4} + e_{1} p_{5} - p_{6}

\Rightarrow p_{6} = - \frac{1}{12} + \frac{1}{2} + 2 + \frac{139}{24} = \frac{197}{24}

g e n e r a l l y

p_{n} = p_{n - 1} + \frac{p_{n - 2}}{2} + \frac{p_{n - 3}}{6} - \frac{p_{n - 4}}{24}

o r

p_{n} = a^{n} + b^{n} + c^{n} + d^{n}

a, b, c, d a r e r o o t s o f e q u a t i o n

x^{4} - x^{3} - \frac{1}{2} x^{2} - \frac{1}{6} x + \frac{1}{24} = 0

Commented by Tawa11 last updated on 23/Jul/21

Thanks sir . God bless you .

Commented by mr W last updated on 23/Jul/21

https://en.m.wikipedia.org/wiki/Newton%27s_identities

Commented by Rasheed.Sindhi last updated on 23/Jul/21

W o n d e r f u l m e t h o d s i r!

B u t I^{'} m f e e l i n g d i f f i c u l i t y i n

u n d e r s t a n d i n g .

Commented by mr W last updated on 23/Jul/21

s e e a l s o Q 74970

Commented by Tawa11 last updated on 23/Jul/21

Sir, which sequence you use to get the P_{n} . Just the sequence .

Commented by Tawa11 last updated on 23/Jul/21

Or there is a format to get the P_{n}

Commented by Tawa11 last updated on 24/Jul/21

I grab now, you used e_{1} - e_{2} + e_{3} - e_{4}

Commented by peter frank last updated on 24/Jul/21

t h a n k y o u