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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Sequences and Series

NDA 2015 · Mathematics · Sequences and Series

The arithmetic mean of 1,8,27,64, . up to n terms is given by

The arithmetic mean of 1,8,27,64,…1,8,27,64, \ldots. up to nn terms is given by

Options

  1. An(n+1)2\frac{\mathrm{n}(\mathrm{n}+1)}{2}
  2. Bn(n+1)22\frac{\mathrm{n}(\mathrm{n}+1)^2}{2}
  3. Cn(n+1)24\frac{\mathrm{n}(\mathrm{n}+1)^2}{4}
  4. Dn2(n+1)24\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}

Correct answer

C. n(n+1)24\frac{\mathrm{n}(\mathrm{n}+1)^2}{4}

Step-by-step solution

Sum of the given terms =1+8+27+64………+=1+8+27+64 \ldots \ldots \ldots+ upto nn terms Sum =13+23+33+43+=1^3+2^3+3^3+4^3+\qquad + upto n terms Sum =(n(n+1)2)2=n2(n+1)24AM=n2(n+1)24×1n=n(n+1)24=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^2=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\mathrm{AM}=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4} \times \frac{1}{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)^2}{4}
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