Question

Prove that, if 9 is added to the sum of first few terms of

the arithmetic sequence: 16, 24, 32, ...........gives a

perfect square?​

Answer 1

Step-by-step explanation:

Step-by-step explanation:

16,24,32,,................

here a=16,d=8

so

Sn=n/2(2*16+(n-1)8)

=n/2(32 +8n-8)

=n/2(8n+24)

Sn=n(4n+12)=4n²+12n

Adding 9 both sides

Sn+9=4n²+12n+9

=(2n)²+2*2n*3+(3)²

Sn+9=(2n+3)²

hence Sn+9 is a perfect square

hence proved

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❤️ jai hind ❤️

Answer 2

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16,24,32,,................

here a=16,d=8

so

Sn=n/2(2*16+(n-1)8)

=n/2(32 +8n-8)

=n/2(8n+24)

Sn=n(4n+12)=4n²+12n

Adding 9 both sides

Sn+9=4n²+12n+9

=(2n)²+2*2n*3+(3)²

Sn+9=(2n+3)²

hence Sn+9 is a perfect square

hence proved