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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 69

Let f(n)=[13+3n100]nf(n)=\left[\dfrac13+\dfrac{3n}{100}\right]n, where [n][n] denotes the greatest integer less than or equal to nn. Then ∑n=156f(n)\displaystyle\sum_{n=1}^{56}f(n) is equal to :

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Answer and worked solution

Official answer: D

[13+3n100]\left[\frac13+\frac{3n}{100}\right] is:

  • 0 for n≤22n\le22;
  • 1 for 23≤n≤5523\le n\le55;
  • 2 for n=56n=56.

∑f(n)=∑2355n+2×56=33(23+55)2+112=1287+112=1399.\sum f(n)=\sum_{23}^{55}n+2\times56=\frac{33(23+55)}2+112=1287+112=1399.

Topics:Special series sumsTypes of functions
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