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 \cdot {2^2} + {3^2} + 2 \cdot {4^2} + {5^2} + 2 \cdot {6^2} + ... $ If B − 2A = 100 λ, then λ is equal to
A. 464
B.496
C. 232
D. 248

Answer 1

Hint: Here, simplify terms of A and B, by writing A and B in some standard series for which formula is known; and by applying formula find the values of A and B. Substitute the values of A and B in B – 2A and compare the result with 100 λ, then value of λ.

Complete step-by-step answer:
 $ A = {1^2} + 2 \cdot {2^2} + {3^2} + 2 \cdot {4^2} + {5^2} + 2 \cdot {6^2} + ... + 2 \cdot {20^2} $
And, $ B = {1^2} + 2 \cdot {2^2} + {3^2} + 2 \cdot {4^2} + {5^2} + 2 \cdot {6^2} + ... + 2 \cdot {40^2} $
Now, A and B can be represented as
\[A = {1^2} + {3^2} + {5^2} + {7^2} + ... + {19^2} + 2\left[ {{2^2} + {4^2} + ... + {{20}^2}} \right]\]
\[B = {1^2} + {3^2} + {5^2} + {7^2} + ... + {39^2} + 2\left[ {{2^2} + {4^2} + ... + {{40}^2}} \right]\]
We know that
Sum of square of first n odd natural number \[ = \dfrac{{n\left( {2n + 1} \right)\left( {2n - 1} \right)}}{3}\]
Sum of square of first n even natural number \[ = \dfrac{{2n\left( {n + 1} \right)\left( {2n + 1} \right)}}{3}\]
Applying formula,
\[A = {\left[ {\dfrac{{n\left( {2n + 1} \right)\left( {2n - 1} \right)}}{3}} \right]} + 2{\left[ {\dfrac{{2n\left( {n + 1} \right)\left( {2n + 1} \right)}}{3}} \right]}\]
 put n=10 and on simplifying,
\[A = {\left[ {\dfrac{{10 \times \left( {2 \times 10 + 1} \right)\left( {2 \times 10 - 1} \right)}}{3}} \right]} + 2\left[ {\dfrac{{2 \times 10 \times \left( {10 + 1} \right)\left( {2 \times 10 + 1} \right)}}{3}} \right]\]
\[ \Rightarrow A = {\left[ {\dfrac{{10 \times 21 \times 19}}{3}} \right]} + 2\left[ {\dfrac{{20 \times 11 \times 21}}{3}} \right] = \dfrac{{13230}}{3}\]
Applying formula
\[B = {\left[ {\dfrac{{n\left( {2n + 1} \right)\left( {2n - 1} \right)}}{3}} \right]} + 2{\left[ {\dfrac{{2n\left( {n + 1} \right)\left( {2n + 1} \right)}}{3}} \right]}\]
 put n=20 and on simplifying,
\[B = {\left[ {\dfrac{{20 \times \left( {2 \times 20 + 1} \right)\left( {2 \times 20 - 1} \right)}}{3}} \right]} + 2\left[ {\dfrac{{2 \times 20 \times \left( {20 + 1} \right)\left( {2 \times 20 + 1} \right)}}{3}} \right]\]
\[ \Rightarrow B = {\left[ {\dfrac{{20 \times 41 \times 39}}{3}} \right]} + 2\left[ {\dfrac{{40 \times 21 \times 41}}{3}} \right] = \dfrac{{100860}}{3}\]
Now, putting values of B and A in B – 2A,
\[B - 2A = \left[ {\dfrac{{100860 - 26460}}{3}} \right] = \dfrac{{74400}}{3}\]
B − 2A = 24800 = 100λ
On comparing both sides, we get
$λ = 248$

Note: In these types of questions, first simplify the expression as expression is not given in either A.P. or G.P. Change the sequence in either A.P. or G.P. or any other standard sequence for which formula is known.
Try to find some pattern and sequence and arrange accordingly. And then solve the expression to get a simplified value. In most of the series asked in question, they belong to AP. or G.P. or some other standard sequence but in mixed form, so first separate the terms and write in proper structure. Do not try to find the values of each term and then arrange in series this can make the sequence more complicated. Therefore do simplify the terms only arrange the terms.