Expand 52 x in descending powers ofx up to the 4th term hence evaluate52.02

To expand (1+x)^n in descending powers of x up to the 4th term, we can use the binomial theorem, which states that:

(1+x)^n = 1 + nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3! + ...

So, for (1+x)^52, the expansion up to the 4th term is:

(1+x)^52 = 1 + 52x + 5251x^2/2! + 5251*50x^3/3!

Simplify the above expression:

= 1 + 52x + 1326x^2 + 22100x^3

Now, to evaluate (1.02)^52 using the above expansion, replace x with 0.02 in the above expression:

= 1 + 52(0.02) + 1326(0.02)^2 + 22100(0.02)^3

= 1 + 1.04 + 0.5292 + 0.169344

= 2.738544

So, (1.02)^52 ≈ 2.738544 when rounded to six decimal places.