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प्रश्न
22 + 42 + 62 + 82 + ...
उत्तर
Let \[T_n\] be the nth term of the given series.
Thus, we have: \[T_n = \left( 2n \right)^2\]
Now, let \[S_n\] be the sum of n terms of the given series.
Thus, we have:
\[S_n = \sum^n_{k = 1} T_k \]
\[ = \sum^n_{k = 1} \left( 2k \right)^2 \]
\[ = \sum^n_{k = 1} 4 k^2 \]
\[ = 4 \sum^n_{k = 1} k^2 \]
\[ = 4\frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6}\]
\[ = \frac{2n}{3}\left( n + 1 \right)\left( 2n + 1 \right)\]
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