Question

Statement – 1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + ...... + (361 + 380 + 400) is 8000.
Statement – 2: $\sum\limits_{k = 1}^n {\left( {{k^3} - {{\left( {k - 1} \right)}^3}} \right)} = {n^3}$, for any natural number n.
$\left( a \right)$ Statement 1 is false, statement 2 is true.
$\left( b \right)$ Statement 1 is true, statement 2 is true. Statement 2 is a correct explanation for statement 1.
$\left( c \right)$ Statement 1 is true, statement 2 is true. Statement 2 is not a correct explanation for statement 1.
$\left( d \right)$ Statement 1 is true, statement 2 is false.

Answer 1

Hint: In this particular question use the concept of expanding the summation, after expanding check which terms are cancel out and which are not, the terms which are not cancel out is the solution for the statement 2, later on in the solution use the concept of $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
 Statement – 2: $\sum\limits_{k = 1}^n {\left( {{k^3} - {{\left( {k - 1} \right)}^3}} \right)} = {n^3}$, for any natural number n.
Consider the LHS of the above equation we have,
$ \Rightarrow \sum\limits_{k = 1}^n {\left( {{k^3} - {{\left( {k - 1} \right)}^3}} \right)} $
Now expand this summation we have,
$ \Rightarrow \left( {{1^3} - {{\left( {1 - 1} \right)}^3}} \right) + \left( {{2^3} - {{\left( {2 - 1} \right)}^3}} \right) + \left( {{3^3} - {{\left( {3 - 1} \right)}^3}} \right) + ...... + \left( {{{\left( {n - 1} \right)}^3} - {{\left( {\left( {n - 1} \right) - 1} \right)}^3}} \right) + \left( {{n^3} - {{\left( {n - 1} \right)}^3}} \right)$....................... (1)
Now simplify this we have,
$ \Rightarrow {1^3} - {\left( 0 \right)^3} + {2^3} - {\left( 1 \right)^3} + {3^3} - {\left( 2 \right)^3} + ...... + {\left( {n - 1} \right)^3} - {\left( {n - 2} \right)^3} + {n^3} - {\left( {n - 1} \right)^3}$
So as we see that all the terms of the above series are canceled except ${n^3}$.
$ \Rightarrow \sum\limits_{k = 1}^n {\left( {{k^3} - {{\left( {k - 1} \right)}^3}} \right)} = {n^3}$
= RHS
So statement 2 is the correct answer.
Statement – 1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + ...... + (361 + 380 + 400) is 8000.
Consider the LHS of the above equation we have,
1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + ...... + (361 + 380 + 400)
Now in equation (1) substitute n = 20 we have
$ \Rightarrow \left( {{1^3} - {{\left( {1 - 1} \right)}^3}} \right) + \left( {{2^3} - {{\left( {2 - 1} \right)}^3}} \right) + \left( {{3^3} - {{\left( {3 - 1} \right)}^3}} \right) + \left( {{4^3} - {{\left( {4 - 1} \right)}^3}} \right) + ...... + \left( {{{\left( {20} \right)}^3} - {{\left( {20 - 1} \right)}^3}} \right)$
And this value is equal to ${n^3}$ according to statement 2 so we have,
$ \Rightarrow \left( {{1^3} - {{\left( {1 - 1} \right)}^3}} \right) + \left( {{2^3} - {{\left( {2 - 1} \right)}^3}} \right) + \left( {{3^3} - {{\left( {3 - 1} \right)}^3}} \right) + \left( {{4^3} - {{\left( {4 - 1} \right)}^3}} \right) + ...... + \left( {{{\left( {20} \right)}^3} - {{\left( {20 - 1} \right)}^3}} \right) = {\left( {20} \right)^3}$
$ \Rightarrow \left( {{1^3} - {{\left( {1 - 1} \right)}^3}} \right) + \left( {{2^3} - {{\left( {2 - 1} \right)}^3}} \right) + \left( {{3^3} - {{\left( {3 - 1} \right)}^3}} \right) + ...... + \left( {{{\left( {20} \right)}^3} - {{\left( {20 - 1} \right)}^3}} \right) = {\left( {20} \right)^3}$
$ \Rightarrow \left( {{1^3} - {{\left( 0 \right)}^3}} \right) + \left( {{2^3} - {{\left( 1 \right)}^3}} \right) + \left( {{3^3} - {{\left( 2 \right)}^3}} \right) + \left( {{4^3} - {{\left( 3 \right)}^3}} \right)...... + \left( {{{\left( {20} \right)}^3} - {{\left( {19} \right)}^3}} \right) = {\left( {20} \right)^3}$
$ \Rightarrow 1 + \left( {{2^3} - {{\left( 1 \right)}^3}} \right) + \left( {{3^3} - {{\left( 2 \right)}^3}} \right) + \left( {{4^3} - {{\left( 3 \right)}^3}} \right)...... + \left( {{{\left( {20} \right)}^3} - {{\left( {19} \right)}^3}} \right) = {\left( {20} \right)^3}$
Now as we know that $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ so in this property in the above equation we have,
$ \Rightarrow 1 + \left( {2 - 1} \right)\left( {{2^2} + {1^2} + 2} \right) + \left( {3 - 2} \right)\left( {{3^2} + {2^2} + 6} \right) + \left( {4 - 3} \right)\left( {{4^2} + {3^2} + 12} \right)...... + \left( {20 - 19} \right)\left( {{{20}^2} + {{19}^2} + 20\left( {19} \right)} \right) = {\left( {20} \right)^3}$
Now simplify we have,
$ \Rightarrow 1 + \left( {4 + 1 + 2} \right) + \left( {9 + 4 + 6} \right) + \left( {16 + 9 + 12} \right)...... + \left( {400 + 361 + 384} \right) = {\left( {20} \right)^3}$
$ \Rightarrow 1 + \left( {1 + 2 + 4} \right) + \left( {4 + 6 + 9} \right) + \left( {9 + 12 + 16} \right)...... + \left( {361 + 384 + 400} \right) = 8000$
= RHS
So statement 1 is also correct, and statement 2 is a correct explanation for statement 1.
Hence option (b) is the correct answer.

Note: In such types of questions it is advised to simplify the LHS or the RHS according to their complexity of given functions. Sometimes proving LHS = RHS needs simplification on both sides of the equation. Remember to convert dissimilar functions to get to the final result, and check whether R.H.S is equal to L.H.S or not if yes then it is the required answer, and also check whether statement 2 is used to simplify the statement 1, if yes then both are true and statement 2 is a correct explanation for statement 1.