Maths · Arithmetic and Geometric progressions
The sum of the following series }{\mathbf{7}}+ \) }{9}+ \) }{11}+\ldots . \) up
The sum of the following series \( 1+6+ \) \( \frac{\mathbf{9}\left(\mathbf{1}^{2}+\mathbf{2}^{2}+\mathbf{3}^{2}\right)}{\mathbf{7}}+ \) \( \frac{12\left(1^{2}+2^{2}+3^{2}+4^{2}\right)}{9}+ \) \( \frac{15\left(1^{2}+2^{2}+\ldots+5^{2}\right)}{11}+\ldots . \) up to 15 terms is:
- A. 7820
- B. 7830
- C. 7520
- D. 7510
Step-by-step solution
The series terms from n=3 simplify to T_n = n^2(n+1)/2. Summing from n=3 to 15 using sum of squares and cubes formulas gives 7813, plus the first two terms 1 and 6 gives 7820.
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