Answer to Question #236801 in Algebra for shoaib

Question

Determine the numerical value of the following expression without the use of a calculator: log10 (1000100) 100 + X 100 n=1 sin(πn) + 1 (−1)n ! · vuut 1000 Y m=1 1 cos(πm) 2 

Answer 1

$(\frac{log_{10}(1000^{100})}{100}+\displaystyle\sum_{n-1}^{100}\frac{sin(πn)+1}{(-1)^n}).\sqrt{\displaystyle\pi_{m-1}^{1000}\frac{1}{cos(πm)^2}}......(1)$

solve

$\frac{log_{10}(1000^{100})}{100}=\frac{100 log_{10}^{1000}}{100}.....[log \space m^{n}=nlog\space m]\\\frac{100log_{10}^{10^{3}}}{100}\\=3log_{10}^{10}\\=3\space......[log_{10}^{10}=1]$

Now

$\displaystyle\sum_{n-1}^{100}\frac{sin(πn)+1}{(-1)^n})=\displaystyle\sum_{n-1}^{100}[\frac{1}{(-1)^n}]=0......(sin\space πn=0)$ and $\sqrt{\displaystyle\pi_{m-1}^{1000}\frac{1}{cos(πm)^2}}=\displaystyle\pi_{m-1}^{1000}\frac{1}{cos(πm)}=1.......(cos\space πm=(-1)^m)$

putting these values in equation 1 we get

$(3+0)\times 1=3$

so

$(\frac{log_{10}(1000^{100})}{100}+\displaystyle\sum_{n-1}^{100}\frac{sin(πn)+1}{(-1)^n}).\sqrt{\displaystyle\pi_{m-1}^{1000}\frac{1}{cos(πm)^2}}=3$