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MATH Solutions By PHANI
605 వీక్షించారు
#🏆పోటీ పరీక్షల స్పెషల్
#➗10th గణితం
#📘ఇంటర్ సైన్స్ & మ్యాథ్స్
#📖ఎడ్యుకేషన్✍
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![🏆పోటీ పరీక్షల స్పెషల్ - Find the sum The series is given by: 32 + 62 3 + 6 3 _ ೧ 3" + 6" = ; 1 4+8 42 + 82 4+8 4" + 8" Solution: Step Factor each term: 32 (]2 _ 22) 33 (13 + 23) 3(1+2) 3" (1"+2") 5 = 42 (12 + 22) 4(1+2) 4" (1" + 2") 43 (13 _ 23) Step 2 Cancel the comnmnon factors: 33 32 = _ 4 4 42 2: Step 3 Multiply equation (1) by 32 4$ = 4 4 3२ 3"+1 5 = 4 4 4"+1 Step 4 Subtract: equation (2) equation (1) 3"+1 $ - 5 [7-[1 3111 3 > 4" Step 5 Solve for S: 3"+1 3 > 4" 4"+1 3"+1 } > 4" _ 3 (4" _ 3 5= $=3 $ Education Reform Centre Find the sum The series is given by: 32 + 62 3 + 6 3 _ ೧ 3" + 6" = ; 1 4+8 42 + 82 4+8 4" + 8" Solution: Step Factor each term: 32 (]2 _ 22) 33 (13 + 23) 3(1+2) 3" (1"+2") 5 = 42 (12 + 22) 4(1+2) 4" (1" + 2") 43 (13 _ 23) Step 2 Cancel the comnmnon factors: 33 32 = _ 4 4 42 2: Step 3 Multiply equation (1) by 32 4$ = 4 4 3२ 3"+1 5 = 4 4 4"+1 Step 4 Subtract: equation (2) equation (1) 3"+1 $ - 5 [7-[1 3111 3 > 4" Step 5 Solve for S: 3"+1 3 > 4" 4"+1 3"+1 } > 4" _ 3 (4" _ 3 5= $=3 $ Education Reform Centre - ShareChat](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_276285_30818dbb_1772616273722_sc.jpg?tenant=sc&referrer=pwa-sharechat-service&f=722_sc.jpg "🏆పోటీ పరీక్షల స్పెషల్ - Find the sum The series is given by: 32 + 62 3 + 6 3 _ ೧ 3\" + 6\" = ; 1 4+8 42 + 82 4+8 4\" + 8\" Solution: Step Factor each term: 32 (]2 _ 22) 33 (13 + 23) 3(1+2) 3\" (1\"+2\") 5 = 42 (12 + 22) 4(1+2) 4\" (1\" + 2\") 43 (13 _ 23) Step 2 Cancel the comnmnon factors: 33 32 = _ 4 4 42 2: Step 3 Multiply equation (1) by 32 4$ = 4 4 3२ 3\"+1 5 = 4 4 4\"+1 Step 4 Subtract: equation (2) equation (1) 3\"+1 $ - 5 [7-[1 3111 3 > 4\" Step 5 Solve for S: 3\"+1 3 > 4\" 4\"+1 3\"+1 } > 4\" _ 3 (4\" _ 3 5= $=3 $ Education Reform Centre Find the sum The series is given by: 32 + 62 3 + 6 3 _ ೧ 3\" + 6\" = ; 1 4+8 42 + 82 4+8 4\" + 8\" Solution: Step Factor each term: 32 (]2 _ 22) 33 (13 + 23) 3(1+2) 3\" (1\"+2\") 5 = 42 (12 + 22) 4(1+2) 4\" (1\" + 2\") 43 (13 _ 23) Step 2 Cancel the comnmnon factors: 33 32 = _ 4 4 42 2: Step 3 Multiply equation (1) by 32 4$ = 4 4 3२ 3\"+1 5 = 4 4 4\"+1 Step 4 Subtract: equation (2) equation (1) 3\"+1 $ - 5 [7-[1 3111 3 > 4\" Step 5 Solve for S: 3\"+1 3 > 4\" 4\"+1 3\"+1 } > 4\" _ 3 (4\" _ 3 5= $=3 $ Education Reform Centre - ShareChat")