312×22+522×32+312×22+522×32+\frac{3}{{{1^2} \times {2^2}}} + \frac{5}{{{2^2} \times {3^2}}} + 732×42+942×52+" role="presentation">732×42+942×52+732×42+942×52+\frac{7}{{{3^2} \times {4^2}}} + \frac{9}{{{4^2} \times {5^2}}} + 1152×62+1362×72+" role="presentation">1152×62+1362×72+1152×62+1362×72+\frac{{11}}{{{5^2} \times {6^2}}} + \frac{{13}}{{{6^2} \times {7^2}}} + 1572×82+1782×92+" role="presentation">1572×82+1782×92+1572×82+1782×92+\frac{{15}}{{{7^2} \times {8^2}}} + \frac{{17}}{{{8^2} \times {9^2}}} + 1992×102is = ?" role="presentation">1992×102is = ?1992×102is = ?\frac{{19}}{{{9^2} \times {{10}^2}}}\,\,\,\,{\text{is = ?}}" /> 312×22+522×32+312×22+522×32+\frac{3}{{{1^2} \times {2^2}}} + \frac{5}{{{2^2} \times {3^2}}} + 732×42+942×52+" role="presentation">732×42+942×52+732×42+942×52+\frac{7}{{{3^2} \times {4^2}}} + \frac{9}{{{4^2} \times {5^2}}} + 1152×62+1362×72+" role="presentation">1152×62+1362×72+1152×62+1362×72+\frac{{11}}{{{5^2} \times {6^2}}} + \frac{{13}}{{{6^2} \times {7^2}}} + 1572×82+1782×92+" role="presentation">1572×82+1782×92+1572×82+1782×92+\frac{{15}}{{{7^2} \times {8^2}}} + \frac{{17}}{{{8^2} \times {9^2}}} + 1992×102is = ?" role="presentation">1992×102is = ?1992×102is = ?\frac{{19}}{{{9^2} \times {{10}^2}}}\,\,\,\,{\text{is = ?}}" /> 312×22+522×32+312×22+522×32+\frac{3}{{{1^2} \times {2^2}}} + \frac{5}{{{2^2} \times {3^2}}} + 732×42+942×52+" role="presentation">732×42+942×52+732×42+942×52+\frac{7}{{{3^2} \times {4^2}}} + \frac{9}{{{4^2} \times {5^2}}} + 1152×62+1362×72+" role="presentation">1152×62+1362×72+1152×62+1362×72+\frac{{11}}{{{5^2} \times {6^2}}} + \frac{{13}}{{{6^2} \times {7^2}}} + 1572×82+1782×92+" role="presentation">1572×82+1782×92+1572×82+1782×92+\frac{{15}}{{{7^2} \times {8^2}}} + \frac{{17}}{{{8^2} \times {9^2}}} + 1992×102is = ?" role="presentation">1992×102is = ?1992×102is = ?\frac{{19}}{{{9^2} \times {{10}^2}}}\,\,\,\,{\text{is = ?}}" />

India’s No.1 Educational Platform For UPSC,PSC And All Competitive Exam

Download Our App

The value of $\frac{3}{1^{2} \times 2^{2}} + \frac{5}{2^{2} \times 3^{2}} +$ $\frac{3}{{{1^2} \times {2^2}}} + \frac{5}{{{2^2} \times {3^2}}} +$ $\frac{7}{3^{2} \times 4^{2}} + \frac{9}{4^{2} \times 5^{2}} +$ $\frac{7}{{{3^2} \times {4^2}}} + \frac{9}{{{4^2} \times {5^2}}} +$ $\frac{11}{5^{2} \times 6^{2}} + \frac{13}{6^{2} \times 7^{2}} +$ $\frac{{11}}{{{5^2} \times {6^2}}} + \frac{{13}}{{{6^2} \times {7^2}}} +$ $\frac{15}{7^{2} \times 8^{2}} + \frac{17}{8^{2} \times 9^{2}} +$ $\frac{{15}}{{{7^2} \times {8^2}}} + \frac{{17}}{{{8^2} \times {9^2}}} +$ $\frac{19}{9^{2} \times 10^{2}} is = ?$ $\frac{{19}}{{{9^2} \times {{10}^2}}}\,\,\,\,{\text{is = ?}}$

1 $\frac{1}{100}$ $\frac{1}{{100}}$
2 $\frac{99}{100}$ $\frac{{99}}{{100}}$
31
4 $\frac{101}{100}$ $\frac{{101}}{{100}}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression, \\ (\frac{1}{1^{2}} - \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}{9^{2}} - \frac{1}{10^{2}}) \\ = (\frac{1}{1^{2}} - \frac{1}{2^{2}}) \\ = (1 - \frac{1}{100}) \\ = \frac{99}{100} \end{aligned}

$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + \left( {\frac{1}{{{4^2}}} - \frac{1}{{{5^2}}}} \right)....\left( {\frac{1}{{{9^2}}} - \frac{1}{{{{10}^2}}}} \right) \cr & = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) \cr & = \left( {1 - \frac{1}{{100}}} \right) \cr & = \frac{{99}}{{100}} \cr}$

Workspace Report

Post your Comments

\frac{7}{3^{2} \times 4^{2}} + \frac{9}{4^{2} \times 5^{2}} +

$\frac{7}{{{3^2} \times {4^2}}} + \frac{9}{{{4^2} \times {5^2}}} +$

\frac{11}{5^{2} \times 6^{2}} + \frac{13}{6^{2} \times 7^{2}} +

$\frac{{11}}{{{5^2} \times {6^2}}} + \frac{{13}}{{{6^2} \times {7^2}}} +$

\frac{15}{7^{2} \times 8^{2}} + \frac{17}{8^{2} \times 9^{2}} +

$\frac{{15}}{{{7^2} \times {8^2}}} + \frac{{17}}{{{8^2} \times {9^2}}} +$

\frac{19}{9^{2} \times 10^{2}} is  =  ?

$\frac{{19}}{{{9^2} \times {{10}^2}}}\,\,\,\,{\text{is = ?}}$ ", "text": "The value of

\frac{3}{1^{2} \times 2^{2}} + \frac{5}{2^{2} \times 3^{2}} +

$\frac{3}{{{1^2} \times {2^2}}} + \frac{5}{{{2^2} \times {3^2}}} +$

\frac{7}{3^{2} \times 4^{2}} + \frac{9}{4^{2} \times 5^{2}} +

$\frac{7}{{{3^2} \times {4^2}}} + \frac{9}{{{4^2} \times {5^2}}} +$

\frac{11}{5^{2} \times 6^{2}} + \frac{13}{6^{2} \times 7^{2}} +

$\frac{{11}}{{{5^2} \times {6^2}}} + \frac{{13}}{{{6^2} \times {7^2}}} +$

\frac{15}{7^{2} \times 8^{2}} + \frac{17}{8^{2} \times 9^{2}} +

$\frac{{15}}{{{7^2} \times {8^2}}} + \frac{{17}}{{{8^2} \times {9^2}}} +$

\frac{19}{9^{2} \times 10^{2}} is  =  ?

$\frac{{19}}{{{9^2} \times {{10}^2}}}\,\,\,\,{\text{is = ?}}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given expression, \\ (\frac{1}{1^{2}} - \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}{9^{2}} - \frac{1}{10^{2}}) \\ = (\frac{1}{1^{2}} - \frac{1}{2^{2}}) \\ = (1 - \frac{1}{100}) \\ = \frac{99}{100} \end{aligned}

$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + \left( {\frac{1}{{{4^2}}} - \frac{1}{{{5^2}}}} \right)....\left( {\frac{1}{{{9^2}}} - \frac{1}{{{{10}^2}}}} \right) \cr & = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) \cr & = \left( {1 - \frac{1}{{100}}} \right) \cr & = \frac{{99}}{{100}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given expression, \\ (\frac{1}{1^{2}} - \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}{9^{2}} - \frac{1}{10^{2}}) \\ = (\frac{1}{1^{2}} - \frac{1}{2^{2}}) \\ = (1 - \frac{1}{100}) \\ = \frac{99}{100} \end{aligned}

$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + \left( {\frac{1}{{{4^2}}} - \frac{1}{{{5^2}}}} \right)....\left( {\frac{1}{{{9^2}}} - \frac{1}{{{{10}^2}}}} \right) \cr & = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) \cr & = \left( {1 - \frac{1}{{100}}} \right) \cr & = \frac{{99}}{{100}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

Test

Classes

E-Book