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Evaluate: 343^๐‘› + 49^๐‘› โ‹… 7^๐‘›+2/(5 โ‹… 7๐‘›)^3 โˆ’ 25 ร— 7^3โข๐‘› - Mathematics

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Question

Evaluate:

`(343^n + 49^n * 7^(n + 2))/((5 * 7^n)^3 - 25 xx 7^(3n))`

Evaluate
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Solution

Given,

`(343^n + 49^n * 7^(n + 2))/((5 * 7^n)^3 - 25 xx 7^(3n))`

We need to simplify the given expression.

Thus, `(343^n + 49^n xx 7^(n + 2))/((5 xx 7^n)^3 - 25 xx 7^(3n))`

⇒ `((7^3)^n + (7^2)^n xx 7^(n + 2))/((5 xx 7^n)^3 - 5^2 xx 7^(3n))`

⇒ `((7)^(3n) + (7)^(2n) xx 7^(n + 2))/(5^3 xx 7^(3n) - 5^2 xx 7^(3n))`  ...[∴ (an)m = anm]

Now, taking out the common term and simplifying the expression by cancelling out the same term we get,

⇒ `((7)^(3n) + (7)^(2n + n + 2))/(5^2 xx 7^(3n) (5 - 1))`  ...[∴ an × am = an + m]

⇒ `((7)^(3n) + (7)^(3n + 2))/(5^2 xx 7^(3n) xx 4)`

⇒ `((7)^(3n) + (7)^(3n) xx 7^2)/(5^2 xx 7^(3n) xx 4)`   ...[∴ an × am = an + m]

⇒ `((7)^(3n) + (7)^(3n) xx 49)/(25 xx 7^(3n) xx 4)`

Taking out the common term and simplifying the expression by cancelling out the same term we get,

= `((7)^(3n)(1 + 49))/(25 xx 7^(3n) xx 4)`

= `50/(25 xx 4)`

= `2/4`

= `1/2`

Hence, the required is `1/2`.

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Chapter 6: Indices - EXERCISE 6 [Page 67]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 6 Indices
EXERCISE 6 | Q 10. (ii) | Page 67
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