Question

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series
${1}^{2} + 2. {2}^{2} + {3}^{2} + 2. {4}^{2} + {5}^{2} + 2. {6}^{2} + ....$
$\rm \: if \: B-2A = 100 \lambda, \: the n \: \lambda \: is \: equal \: to \:$
​

Answer 1

Answer:

248

Step-by-step explanation:

A=1+2.2^2+32+2.4^2+⋯+20.202  

B=1+2.22+32+2.42+⋯+40.402

For A,

We can split series into two parts

A=12+22+22+32+4242+cdots+202+202

A=12+22+32+42+cdots+202+22+42+62+cdots+202

General formula for sum of squares of first n natural numbers is

S=n(n+1)(2n+1)6

A=20(20+1)(2(20)+1)6+22(12+22+32+42+cdots+102)

A=20(20+1)(2(20)+1)6+22(10(10+1)(2(10)+1)6)

A=4410

Similarity we can find value for B also

B=33620

B−2A=24800

Therefore, λ=248

Answer 2

$\underline{ \underline{ \sf \red{Question}}} \red:$

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series

${1}^{2} + 2. {2}^{2} + {3}^{2} + 2. {4}^{2} + {5}^{2} + 2. {6}^{2} + ....$

$\rm \: if \: B-2A = 100 \lambda, \: the n \: \lambda \: is \: equal \: to \:$

$\underline{ \underline{ \sf \red{Answer}}} \red:$

$\rm{We \: have,}$

$\rm {1}^{2} + 2. {2}^{2} + {3}^{2} + 2. {4}^{2} + {5}^{2} + 2. {6}^{2} + ....$

$\rm \: A = \: sum \: of \: first \: 20 \: terms$

$\rm B = sum \: of \: first \: 40 \: terms$

$\rm \therefore \: A = {1}^{2} + 2. {2}^{2} + {3}^{2} + 2. {4}^{2} + {5}^{2} + 2. {6}^{2} + .... + 2. {20}^{2}$

$\rm \: A = ( {1}^{2} + {2}^{2} + {3}^{2} + ... + {20}^{2} ) + ( {2}^{2} + {4}^{2} + {6}^{2} + ... + {20}^{2} )$

$\rm \: A = ( {1}^{2} + {2}^{2} + {3}^{2} + ... + {20}^{2} ) + 4( {1}^{2} + {2}^{2} + {3}^{2} + ... + {10}^{2} )$

$\rm \: A = \dfrac{20 \times 21 \times 41}{6} + \dfrac{4 \times 10 \times 11 \times 21}{6}$

$\rm \: A \: = \dfrac{20 \times 21}{6}(41 + 22)$

$\boxed{ \boxed{\rm \: A \: = \dfrac{20 \times 41 \times 63}{6} }} - - - (1)$

$\sf \red{Similarly}$

$\rm \: B = ( {1}^{2} + {2}^{2} + {3}^{2} + ... + {40}^{2} ) + 4( {1}^{2} + {2}^{2} + ... + {20}^{2} )$

$\rm \: B \: = \dfrac{40 \times 41 \times 81}{6} + \dfrac{4 \times 20 \times 21 \times 41}{6}$

$\rm \: B = \dfrac{40 \times 41}{6} (81 + 42)$

$\boxed{ \boxed{ \rm \: B = \dfrac{40 \times 41 \times 123}{6} }} - - - (2)$

$\sf{Now ; \: \red{B-2A = 100 \lambda}}$

$\therefore \: \dfrac{40 \times 41 \times 123}{6} - \dfrac{2 \times 20 \times 21 \times 63}{6} = 100 \lambda$

$\implies \: \dfrac{40}{6} (5043 - 1323) = 100 \lambda$

$\implies \dfrac{40}{6} \times 3720 = 100 \lambda$

$\implies \: 40 \times 620 = 100 \lambda$

$\implies \: \lambda \: = \dfrac{40 \times 620}{100}$

$\boxed{ \boxed{ \lambda \: = \: 248}}$

$\underline{\underline{\sf\red{Formula \: Used}}} \red{:}$

$\rm \: Squares \: of \:  first  \: 'n' \: natural  \: numbers \: \implies \\ \\ \rm \boxed{ \boxed{ \sum \: {n}^{2} = \dfrac{n(n + 1)(2n + 1)}{6} }}$