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Proof by mathematical induction

Let A be the ...

Question

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series. $1^{2} + 2 . 2^{2} + 3^{2} + 2 . 4^{2} + 5^{2} + 2 . 6^{2} + . . .$ If $B - 2 A = 100 λ$ , then $λ$ is equal to :

$464$

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$496$

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$232$

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$248$

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Solution

The correct option is D
$248$

Determine the value of $λ$
$A$ be the sum of the first $20$ terms
$A = 1^{2} + 2 . 2^{2} + 3^{2} + 2 . 4^{2} + 5^{2} + 2 . 6^{2} + . . . + 2 . {(20)}^{2} = (1^{2} + 3^{2} + . . . + 19^{2}) + (2 . 2^{2} + 2 . 4^{2} + . . . + 2 . {(20)}^{2}) [seperateoddandeventerm] = (1^{2} + 3^{2} + . . . + 19^{2}) + 2 . (2^{2} + 4^{2} + . . . + {(20)}^{2}) = (1^{2} + 3^{2} + . . . + 19^{2}) + (2^{2} + 4^{2} + . . . + {(20)}^{2}) + (2^{2} + 4^{2} + . . . + {(20)}^{2}) = (1^{2} + 2^{2} + 3^{2} + . . . + {(20)}^{2}) + (2^{2} . 1 + 2^{2} . 2^{2} + . . . + 2^{2} . {(10)}^{2}) = (1^{2} + 2^{2} + 3^{2} + . . . + {(20)}^{2}) + 2^{2} (1^{2} + 2^{2} + 3^{2} + . . . + {(10)}^{2}) = \frac{20 \times 21 \times 41}{6} + 4 [\frac{10 \times 11 \times 21}{6}] [∵ 1^{2} + 2^{2} + 3^{2} + . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}] = \frac{17220}{6} + \frac{4 \times 2310}{6} = 4410$
$B$ is the sum of first $40$ terms.
$B = 1^{2} + 2 . 2^{2} + 3^{2} + 2 . 4^{2} + 5^{2} + 2 . 6^{2} + . . . + 2 . {(40)}^{2} = (1^{2} + 3^{2} + . . . + 39^{2}) + (2 . 2^{2} + 2 . 4^{2} + . . . + 2 . {(40)}^{2}) [seperateoddandeventerm] = (1^{2} + 3^{2} + . . . + 39^{2}) + 2 . (2^{2} + 4^{2} + . . . + {(40)}^{2}) = (1^{2} + 3^{2} + . . . + 39^{2}) + (2^{2} + 4^{2} + . . . + {(40)}^{2}) + (2^{2} + 4^{2} + . . . + {(40)}^{2}) = (1^{2} + 2^{2} + 3^{2} + . . . + {(40)}^{2}) + (2^{2} . 1 + 2^{2} . 2^{2} + . . . + 2^{2} . {(20)}^{2}) = (1^{2} + 2^{2} + 3^{2} + . . . + {(40)}^{2}) + 2^{2} (1^{2} + 2^{2} + 3^{2} + . . . + {(20)}^{2}) = \frac{40 \times 41 \times 81}{6} + 4 [\frac{20 \times 21 \times 41}{6}] [∵ 1^{2} + 2^{2} + 3^{2} + . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}] = \frac{132840}{6} + \frac{4 \times 17220}{6} = 33620$
We know that, $B - 2 A = 100 λ$
therefore, substituting the values we have,
$\Rightarrow 33620 - 2 (4410) = 100 λ \Rightarrow 24800 = 100 λ \Rightarrow λ = 248$
Hence, option (D) is the correct answer

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