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Basic of Logarithms
easy
જો ${1 \over {x(x + 1)\,(x + 2)....(x + n)}} = {{{A_0}} \over x} + {{{A_1}} \over {x + 1}} + {{{A_2}} \over {x + 2}} + .... + {{{A_n}} \over {x + n}}$ તો ${A_r} = $
A
${{r!{{( - 1)}^r}} \over {(n - r)!}}$
B
${{{{( - 1)}^r}} \over {r!(n - r)!}}$
C
${1 \over {r!(n - r)!}}$
D
એકપણ નહીં
Solution
(b) $1 = {A_0}(x + 1)\,(x + 2)….(x + n) + {A_1}x(x + 2)\,(x + 3)…(x + n)$
$ + … + {A_r}x(x + 1)\,(x + 2)….(x + r – 1)\,(x + r + 1)\,(x + r + 2)$ $…….(x + n)$
Putting $x = – r$,
$1 = {A_r}( – r)\,( – r + 1)\,( – r + 2),\,…..( – 1).1.2….( – r + n)$
$ \Rightarrow $ $1 = {A_r}.{( – 1)^r}r!.(n – r)\,!$; ${A_r} = {{{{( – 1)}^r}} \over {r\,!\,(n – r)\,!}}$.
Standard 11
Mathematics
