MHT CET · Maths · Sequences and Series
The sum to 10 terms of the series \(1 \times 3^{2}+2 \times 5^{2}+3 \times 7^{2}+\ldots \ldots \ldots \ldots . .\) is
- A 13,495
- B \(15,595\)
- C \(13,000\)
- D 13,695
Answer & Solution
Correct Answer
(D) 13,695
Step-by-step Solution
Detailed explanation
\(1 \times 3^{2}+2 \times 5^{2}+3 \times 7^{2}+\)
Term : \(\rightarrow 2(2 q+1)^{2}\)
\(S=\sum_{\varepsilon=1}^{10} z(2 q+1)^{2}\)
\(S=\sum_{\varepsilon=1}^{10}\left(4 z^{2}+4 z+1\right) \varepsilon\)
\(S=\sum_{z=1}^{10}\left[4 z^{3}+4 z^{2}+2\right]\)
\(S=\frac{4 \times 10^{2} \times 11^{2}}{4}+\frac{4 \times 10 \times 11 x}{6}$$+\frac{10 \times 11}{2}\)
\(S=13,695\)
Term : \(\rightarrow 2(2 q+1)^{2}\)
\(S=\sum_{\varepsilon=1}^{10} z(2 q+1)^{2}\)
\(S=\sum_{\varepsilon=1}^{10}\left(4 z^{2}+4 z+1\right) \varepsilon\)
\(S=\sum_{z=1}^{10}\left[4 z^{3}+4 z^{2}+2\right]\)
\(S=\frac{4 \times 10^{2} \times 11^{2}}{4}+\frac{4 \times 10 \times 11 x}{6}$$+\frac{10 \times 11}{2}\)
\(S=13,695\)
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