If a+b+c+d=0, shew that 4a5+b5+c5+d5=3a3+b3+c3+d3⋅2a2+b2+c2+d2.

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If a + b + c + d = 0, shew that {a^{5} + b^{5} + c^{5} + d^{5

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If $a + b + c + d = 0$ , shew that $\frac{a ^{5} + b ^{5} + c ^{5} + d ^{5}}{4} = \frac{a ^{3} + b ^{3} + c ^{3} + d ^{3}}{3} \cdot \frac{a ^{2} + b ^{2} + c ^{2} + d ^{2}}{2}$ .

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a + b + c + d = 0

, shew that

\frac{a ^{5} + b ^{5} + c ^{5} + d ^{5}}{4} = \frac{a ^{3} + b ^{3} + c ^{3} + d ^{3}}{3} \cdot \frac{a ^{2} + b ^{2} + c ^{2} + d ^{2}}{2}

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9 mins ago

(1 + a x) (1 + b x) (1 + c x) (1 + d x) = 1 + q x^{2} + r x^{3} + s x^{4} ∵ \sum a = 0

q = \sum ab, r = \sum ab c, s = ab c d

Taking the logarithms and equating the co efficients of

x^{n}

, we have

\frac{( - 1 ) ^{n - 1}}{n} (a^{n} + b^{n} + c^{n} + d^{n})

=

co efficient of

x^{n}

in the expansion of

lo g (1 + q x^{2} + r x^{3} + s x^{4})

=

co efficient of

x^{n}

(q x^{2} + r x^{3} + s x^{4})

= - \frac{1}{2} (q x^{2} + r x^{3} + s x^{4})^{2} + ....

By putting

n = 2, 3, 4, 5

we obtain

- \frac{a ^{2} + b ^{2} + c ^{2} + d ^{2}}{2} = q

\frac{a ^{3} + b ^{3} + c ^{3} + d ^{3}}{3} = r

\frac{a ^{5} + b ^{5} + c ^{5} + d ^{5}}{5} = - q r

∴ \frac{\sum a ^{5}}{5} = \frac{\sum a ^{3}}{3} . \frac{\sum a ^{2}}{2}

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Question Text	If $a + b + c + d = 0$ , shew that $\frac{a ^{5} + b ^{5} + c ^{5} + d ^{5}}{4} = \frac{a ^{3} + b ^{3} + c ^{3} + d ^{3}}{3} \cdot \frac{a ^{2} + b ^{2} + c ^{2} + d ^{2}}{2}$ .
Answer Type	Text solution:1
Upvotes	150