Question

NTSE question try to solve it​

Answer 1

Answer:

2

Step-by-step explanation:

1/(1+√2)*(√2-1)/(√2-1) + 1/(√2+√3)*(√3-√2)/(√3-√2) + .....1/(√8+√9)*(√9-√8)/(√9-√8)

rationalising each term by multiplying them,with its conjugate,

we get,

√2-1 + √3-√2.......+√8-√7+√9-√8

-1+√9

2

no thanks

brainlist the second answer!!!

Answer 2

Question:

Find the value of the series;

1/(1+√2) + 1/(√2+√3) + 1/(√3+√4)

+ 1/(√4+√5) + 1/(√5+√6) + 1/(√6+√7)

+ 1/(√7+√8) + 1/(√8-√9)

Answer:

The value of the given series is 2 .

Solution:

Let , the given series be;

S = 1/(1+√2) + 1/(√2+√3) + 1/(√3+√4)

+ 1/(√4+√5) + 1/(√5+√6) + 1/(√6+√7)

+ 1/(√7+√8) + 1/(√8-√9)

Now,

Rationalizing the denominators of each terms of the series, we get;

=> S = (1-√2)/(-1) + (√2-√3)/(-1)

+ (√3-√4)/(-1) + (√4-√5)/(-1)

+ (√5-√6)/(-1) + (√6-√7)/(-1)

+ (√7-√8)/(-1) + (√8-√9)/(-1)

=> S = -1 + √2 - √2 + √3 - √3 + √4 - √4

+ √5 - √5 + √6 - √6 + √7 - √7

+ √8 - √8 + √9

=> S = -1 + √9

=> S = -1 + 3

=> S = 2

Hence, the required value of the given series is 2 .