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# algebra

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If (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280, then find 1 + 2 + 3 + ... + n.

May 20, 2020

#1
+21953
+1

(2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47)  =  5280

There are 24 terms in the list:   (2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47)

So, the left-hand side can be rewritten as  (2n)(24) + (1 + 3 + 5 + ... + 47)  =  5280

We can use the formula:  Sum  =  n(t1 + tn) / 2 to find the sum of  1 + 3 + ... + 47:

Sum  =  24(1 + 47)/2  =  24(48)/2  =  576

Equation:  (2n)(24) + (576)  =  5280

48n + 576  =  5280

48n  =  4704

n  =  98

Now, to find the sum:  1 + 2 + 3 + ... + 98  using the formula:  Sum  =  n(t1 + tn) / 2

Sum  =  98(1 + 98)/2  =  4851.

May 20, 2020

#1
+21953
+1

(2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47)  =  5280

There are 24 terms in the list:   (2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47)

So, the left-hand side can be rewritten as  (2n)(24) + (1 + 3 + 5 + ... + 47)  =  5280

We can use the formula:  Sum  =  n(t1 + tn) / 2 to find the sum of  1 + 3 + ... + 47:

Sum  =  24(1 + 47)/2  =  24(48)/2  =  576

Equation:  (2n)(24) + (576)  =  5280

48n + 576  =  5280

48n  =  4704

n  =  98

Now, to find the sum:  1 + 2 + 3 + ... + 98  using the formula:  Sum  =  n(t1 + tn) / 2

Sum  =  98(1 + 98)/2  =  4851.

geno3141 May 20, 2020