Question

Please find the answer step by step explanation.....

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Find the value of :- 3/(1²*2²) + 5/(2²*3²) + 7/(3²*4²) + 9/(4²*5²) + 11/(5²*6²) + 13/(6²*7²) + 15/(7²*8²) + 17/(8²*9²) + 19/(9²*10²) ?

Sᴏʟᴜᴛɪᴏɴ :-

Lets First See All Numerator Parts can be written As :-

→ 3 = (4 - 1) = (2² - 1²)

→ 5 = (9 - 4) = (3² - 2²)

→ 7 = (16 - 9) = (4² - 3²)

→ 9 = (25 - 16) = (5² - 4²)

→ 11 = (36 - 25) = (6² - 5²)

→ 13 = (49 - 36) = (7² - 6²)

→ 15 = (64 - 49) = (8² - 7²)

→ 17 = (81 - 64) = (9² - 8²)

→ 19 = (100 - 81) = (10² - 9²)

Putting These values in Numerator now , we get :-

→ (2² - 1²)/(1²*2²) + (3² - 2²)/(2²*3²) + (4² - 3²)/(3²*4²) + (5² - 4²)/(4²*5²) + (6² - 5²)/(5²*6²) + (7² - 6²)/(6²*7²) + (8² - 7²)/(7²*8²) + (9² - 8²)/(8²*9²) + (10² - 9²)/(9²*10²)

Dividing The Numerator with Denominator Now, we get :-

→ (1 - 1/2²) + (1/2² - 1/3²) + (1/3² - 1/4²) + (1/4² - 1/5²) + (1/5² - 1/6²) + (1/6² - 1/7²) + (1/7² - 1/8²) + (1/8² - 1/9²) + (1/9² - 1/10²)

As we can see Except First Term & Last Term All will be cancel ,

So,

→ (1 - 1/10²)

→ (1 - 1/100)

→ (100 - 1)/100

→ (99/100) (B) (Ans.)

(Nice Question).

Answer 2

To find the value of :-

$\displaystyle{\sf{\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2} +\frac{9}{4^2.5^2}+\frac{11}{5^2.6^2}+\frac{13}{6^2.7^2}+\frac{15}{7^2.8^2}+\frac{17}{8^2.9^2} +\frac{19}{9^2.10^2} }}$

Solution :-

We can write the numerators as follows :

• 3 = 4 - 1 = 2² - 1²

• 5 = 9 - 4 = 3² - 2²

• 7 = 16 - 9 = 4² - 3²

• 9 = 25 - 16 = 5² - 4²

• 11 = 36 - 25 = 6² - 5²

• 13 = 49 - 36 = 7² - 6²

• 15 = 64 - 49 = 8² - 7²

• 17 = 81 - 64 = 9² - 8²

• 19 = 100 - 81 = 10² - 9²

Now, substituting these values :

$\displaystyle{\sf{\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2} +\frac{5^2-4^2}{4^2.5^2}+\frac{6^2-5^2}{5^2.6^2}+\frac{7^2-6^2}{6^2.7^2}+\frac{8^2-7^2}{7^2.8^2} }}\\\\\displaystyle{\sf{+\frac{9^2-8^2}{8^2.9^2}+\frac{10^2-9^2}{9^2.10^2} }}$

$\displaystyle{\sf{=(1-\frac{1}{2^2})+(\frac{1}{2^2}-\frac{1}{3^2} )+(\frac{1}{3^2}-\frac{1}{4^2})+(\frac{1}{4^2}-\frac{1}{5^2})+(\frac{1}{5^2} -\frac{1}{6^2} )+(\frac{1}{6^2}-\frac{1}{7^2} ) }}\\\\\displaystyle{\sf{+(\frac{1}{7^2} -\frac{1}{8^2} )+(\frac{1}{8^2} -\frac{1}{9^2} )+(\frac{1}{9^2} -\frac{1}{10^2} ) }}$

When we open the parentheses, we will find that maximum of the terms cancel each other. And, finally we get :

$\displaystyle{\sf{1-\frac{1}{10^2} }}\\\\\\\displaystyle{\sf{ =1-\frac{1}{100} }}\\\\\\\displaystyle{\sf{ =\frac{100-1}{100} }}\\\\\\\huge\underline{\boxed{\boxed{\displaystyle{\sf{ =\frac{99}{100} }}}}}$

∴ So, the answer is (B) 99/100.