Solve 3*(1+3)^2+2^2+2^3 | Microsoft Math Solver

Evaluate

60

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Solution Steps

3 \cdot (1 + 3)^{2} + 2^{2} + 2^{3}

Add

1

and

3

to get

4

.

3 \times 4^{2} + 2^{2} + 2^{3}

Calculate

4

to the power of

2

and get

16

.

3 \times 16 + 2^{2} + 2^{3}

Multiply

3

and

16

to get

48

.

48 + 2^{2} + 2^{3}

Calculate

2

to the power of

2

and get

4

.

48 + 4 + 2^{3}

Add

48

and

4

to get

52

.

52 + 2^{3}

Calculate

2

to the power of

3

and get

8

.

52 + 8

Add

52

and

8

to get

60

.

60

Factor

2^{2} \times 3 \times 5

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Quiz

5 problems similar to:

3 \cdot (1 + 3)^{2} + 2^{2} + 2^{3}

Similar Problems from Web Search

“Counting Tricks”: using combination to derive a general formula for

1^{2} + 2^{2} + \dots + n^{2}

https://math.stackexchange.com/q/47509

The underlying mechanism is known as the calculus of finite differences . Just like one can reconstruct a

n

th degree polynomial

f

f (a), f^{'} (a), \dots, f^{(n)} (a)

a

k^{2}

1^{2} + 2^{2} + \dots + (k - 1)^{2}

https://math.stackexchange.com/questions/2736190/adding-k2-to-12-22-cdots-k-12

The reason why,

1^{2} + 2^{2} + \dots + (k - 1)^{2},

1^{2} + 2^{2} + \dots + (k - 1)^{2} + k^{2},

is because you're adding the

k

th term squared on both sides. Thus, instead of having the sum up to ...

How do you insert parentheses to make

18 + 2 \cdot 1 + 3^{2} = 38

https://socratic.org/questions/how-do-you-insert-parentheses-to-make-18-2-1-3-2-38-true

18 + 2 \cdot (1 + 3^{2})

18 + 2 \cdot (1 + 3^{2}) = 18 + 2 \cdot (1 + 9) = 18 + 2 \cdot 10 = 18 + 20 = 38

Show group of order

4563 = 3^{3} \cdot 1 3^{2}

https://math.stackexchange.com/questions/2951028/show-group-of-order-4563-33-cdot132-is-not-simple

The next things to try are often: to consider the intersections of Sylow groups, and to use the fact that groups of order

p^{2}

p

a prime, are always abelian. Let's see. Assume that we have two ...

Is there a name for the sum of increasing powers?

https://math.stackexchange.com/q/1892037

This is called a geometric sum/series according as

N

is finite or not. More generally, it is of the form

\sum_{n = 0}^{N} a r^{n}

N

may be "equal to" infinity. In the case where

N

Prove that there is no subgroup of index

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in a simple group of order

240

https://math.stackexchange.com/q/2607332

φ : G \to S_{6}

be the permutation representation on the six cosets. This is an embedding since

G

is simple, so we identify

G

with a subgroup of

S_{6}

. Simplicity implies all elements of ...

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3\times 4^{2}+2^{2}+2^{3}

Add 1 and 3 to get 4.

3\times 16+2^{2}+2^{3}

Calculate 4 to the power of 2 and get 16.

48+2^{2}+2^{3}

Multiply 3 and 16 to get 48.

48+4+2^{3}

Calculate 2 to the power of 2 and get 4.

52+2^{3}

Add 48 and 4 to get 52.

52+8

Calculate 2 to the power of 3 and get 8.

60

Add 52 and 8 to get 60.

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