Evaluate - a2n-1- a2n-12n-1-an4n-1-a22n-3

A^2n+1× a^(2n+1)(2n 1)/a^n(4n 1)×(a^2)^2n+3 =a^2n+1× a^(2n)^2 (1)^2/a^4n^2 n× a^2(2n+3) =a^2n+1× a^4n^2 1/a^4n^2 n× a^4n+6 =a^2n+1+4n^2 1 4n^2+n 4n 6 =a^ n ...

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Byju's Answer

Standard VIII

Mathematics

Exponents with like Bases

Evaluate : a2...

Question

Evaluate :

$\frac{a^{2 n + 1} \times a^{(2 n + 1) (2 n - 1)}}{a^{n (4 n - 1)} \times (a^{2})^{2 n + 3}}$

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Solution

$\frac{a^{2 n + 1} \times a^{(2 n + 1) (2 n - 1)}}{a^{n (4 n - 1)} \times (a^{2})^{2 n + 3}} = \frac{a^{2 n + 1} \times a^{(2 n)^{2} - (1)^{2}}}{a^{4 n^{2} - n} \times a^{2 (2 n + 3)}} = \frac{a^{2 n + 1} \times a^{4 n^{2} - 1}}{a^{4 n^{2} - n} \times a^{4 n + 6}} = a^{2 n + 1 + 4 n^{2} - 1 - 4 n^{2} + n - 4 n - 6} = a^{- n - 6} = a^{- (n + 6)} = \frac{1}{a^{n + 6}}$

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Similar questions

Q. State True or False.

Evaluate:

\frac{a^{2 n + 1} \times a^{(2 n + 1) (2 n - 1)}}{a^{n (4 n - 1)} \times (a^{2})^{2 n + 3}}

, then answer is

\frac{1}{a^{n + 6}}

Q. Let the sequence

a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}

form an AP. Then,

a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - . . . . + a_{2 n - 1}^{2} - a_{2 n}^{2}

is?

Q. If

a_{1,} a_{2,} a_{3,} . . . . . . . . a_{2 n + 1}

are in A.P. then

\frac{a_{2 n + 1} - a_{1}}{a_{2 n + 1} + a_{1}} + \frac{a_{2 n + 1} - a_{2}}{a_{2 n + 1} + a_{2}} + . . . . . . + \frac{a_{2 n + 1} - a_{n}}{a_{2 n + 1} + a_{n}}

Q. The standard deviation of the data:

x:	1	a	a²	....	aⁿ
f:	ⁿC₀	ⁿC₁	ⁿC₂	....	ⁿC_n

is
(a)

{(\frac{1 + a^{2}}{2})}^{n} - {(\frac{1 + a}{2})}^{n}

(b)

{(\frac{1 + a^{2}}{2})}^{2 n} - {(\frac{1 + a}{2})}^{n}

(c)

{(\frac{1 + a}{2})}^{2 n} - {(\frac{1 + a^{2}}{2})}^{n}

(d) none of these

Q. If the sequence

a_{1}, a_{2}, a_{3}, . . . . ., a_{n}

. forms an A.P., then prove that

a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + . . . . . + a_{2 n - 1}^{2} - a_{2 n}^{2} = \frac{n}{2 n - 1} (a_{1}^{2} - a_{2 n}^{2})

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Evaluate : a2n+1×a(2n+1)(2n−1)an(4n−1)×(a2)2n+3

a2n+1×a(2n+1)(2n−1)an(4n−1)×(a2)2n+3=a2n+1×a(2n)2−(1)2a4n2−n×a2(2n+3)=a2n+1×a4n2−1a4n2−n×a4n+6=a2n+1+4n2−1−4n2+n−4n−6=a−n−6=a−(n+6)=1an+6

Evaluate :

$\frac{a^{2 n + 1} \times a^{(2 n + 1) (2 n - 1)}}{a^{n (4 n - 1)} \times (a^{2})^{2 n + 3}}$