Question

the period of sin (x+8x+27x+....+n3x)is ​

Answer 1

Question :- The period of sin(x + 8x + 27x+....+n³x) is ?

Solution :-

As we know the property of period, that :-

• if f(x) is function whose period is T.

Than,

• Period of f(ax) will be = T / |a| . ( where a is constant.)

we also know that period of sinx is 2π.

So,

By property of period ,

→ sinx = 2π

Than,

→ sin(ax) = 2π / |a|

Now, we have to find period of sin(x + 8x + 27x+....+n³x) ?

→ sin(x + 8x + 27x+....+n³x)

Taking x common,

→ sin{x(1 + 8 + 27 + _______ n³)}

→ sin{x(1³ + 2³ + 3³ + _______ n³)}

• sum of first n natural numbers cube = {n(n+1)/2}² = {n(n+1)}² / 4

so, period of :-

→ sin[x{n²(n+1)² / 4}]

here ,

• a = {n²(n+1)² / 4} (constant term.)

Therefore,

→ Required period of given function = 2π / constant term.

→ Required period = 2π / {n²(n+1)² / 4}

→ Required period = 2π * 4 / n²(n+1)²

→ Required period = 8π / n²(n +1)² . (Ans.)