Question

If P ( n ) : 1^3 + 2^3 +3^3 + ….. + ( n + 1)^3 = k, then L.H.S. of P ( 2 ) =

Answer 1

Answer:

p(2)=13+23+33+(n+1)3=k

=36+33+3n+3=k

=69+3+3n=k

=72+3×2=k

=72+6=k

=78=k

-78=k

Answer 2

$\huge{\mathbb{\red{ANSWER:-}}}$

Given :-

$\sf{p(n) = 1^{3} + 2^{3} + 3^{3} + ......+ (n+1)^{3} = k}$

To Find :-

$\sf{L.H.S \: of \: p(2) = ? }$

Using Formula :-

$\sf{Sum \: of \: the \: cubes \: of \: first \: n \: natural \: numbers-}$

$\sf{[\dfrac{n(n+1)}{2}]^{2}}$

Solution :-

$\sf{L . H . S :-}$

$\sf{p(n) = 1^{3} + 2^{3} + ......+ (n+1)^{3}}$

$\sf{p(n) =[\dfrac{(n+1)(n+2)}{2}]^{2}}$

$\sf{p(2) =[\dfrac{(2+1)(2+2)}{2}]^{2}}$

$\sf{p(2) =[\dfrac{3\times 4}{2}]^{2}}$

$\sf{p(2) =(3\times 2)^{2}}$

$\sf{p(2) = 6^{2}}$

$\sf{p(2) = 36}$

$\sf{then \: ,}$

$\sf{k = 36}$

Result :-

$\sf{L.H.S \: of \: p(2) = 36}$

Extra Related Formulas :-

1)$\sf{1 + 2 + 3 +......+ n =\dfrac{n(n+1)}{2}}$

2)$\sf{1^{2} + 2^{2} + 3^{2} +......+ n^{2}=\dfrac{n(n+1)(2n+1)}{6}}$