What is expand (t+Δt)^4 ?

The solution to expand (t+Δt)^4 is t^4+4Δt^4+6Δ^2t^4+4Δ^3t^4+Δ^4t^4

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expand (t+Δt)^4

expand

(t + Δ t)^{4}

t^{4} + 4Δ t^{4} + 6 Δ^{2} t^{4} + 4 Δ^{3} t^{4} + Δ^{4} t^{4}

Solution steps

(t + Δ t)^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = t, b = Δ t

= i = 0 \sum 4 (i 4) t^{(4 - i)} (Δ t)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} t^{4} (Δ t)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} t^{3} (Δ t)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} t^{2} (Δ t)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} t^{1} (Δ t)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} t^{0} (Δ t)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} t^{4} (Δ t)^{0} : t^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} t^{3} (Δ t)^{1} : 4Δ t^{4}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} t^{2} (Δ t)^{2} : 6 Δ^{2} t^{4}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} t^{1} (Δ t)^{3} : 4 Δ^{3} t^{4}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} t^{0} (Δ t)^{4} : Δ^{4} t^{4}

= t^{4} + 4Δ t^{4} + 6 Δ^{2} t^{4} + 4 Δ^{3} t^{4} + Δ^{4} t^{4}

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What is expand (t+Δt)^4 ?
The solution to expand (t+Δt)^4 is t^4+4Δt^4+6Δ^2t^4+4Δ^3t^4+Δ^4t^4