- Home
- Standard 12
- Mathematics
माना $f(n)=\left[\frac{1}{3}+\frac{3 n}{100}\right] n$, जहाँ $[n]$ एक महत्तम पूणांक, जो $n$ से छोटा अथवा बराबर है, तो $\sum_{ n =1}^{56} f(u)$ बराबर है
$56$
$689$
$1287$
$1399$
Solution
Let $f\left( n \right) = \left[ {\frac{1}{3} + \frac{{3n}}{{100}}} \right]n$
where $\left[ n \right]$ is greatest integer functon,
$ = \left[ {0.33 + \frac{{3n}}{{100}}} \right]n$
For $n = 1,2,….,22,$ we get $f\left( n \right) = 0$
and for $n = 23,24,….,55,$ we get $f\left( n \right) = 1$
For $n = 56,f\left( n \right) = 2$
So, $\sum\limits_{n = 1}^{56} {f\left( n \right) = 1\left( {23} \right) + 1\left( {24} \right) + … + 1\left( {55} \right)} + 2\left( {56} \right)$
$ = \left( {23 + 24 + ….. + 55} \right) + 112$
$ = \frac{{33}}{2}\left[ {46 + 32} \right] + 112$
$ = \frac{{33}}{2}\left( {78} \right) + 112 = 1399$.
