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Question

If 112+122+132 upto =π26, then 112+132+152+=

A
π212
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B
π224
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C
π28
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D
none of these
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Solution

The correct option is C π28
112+122+132+=π26

(112+132+)+(122+142+)=π26

(112+132+)+122(112+122+)=π26

(112+132+)+14×π26=π26

(112+132+)=π26π224=3π224=π28

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